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Tricks & Tips Of Three Dimensional Geometry For NDA

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Distance between two points and section formula

Q 1619345210

Let `alpha , beta` and `gamma` be distinct real numbers., The points
with position vectors `alpha hat i + beta hat j + gamma hat k, beta hat i + gamma hat j + alpha hat k`
and `gamma hat i + alpha hat j + beta hat k`
NDA Paper 1 2015
(A)

are collinear

(B)

form an equilateral triangle

(C)

form a scalene triangle

(D)

form a right-angled triangle

Solution:

Let `A, B` and `C` be the given points whose

position vectors are `alpha hat i + beta hat j + gamma hat k, beta hat i + gamma hat j + alpha hat k`

and `gamma hat i + alpha hat j + beta hat k` respectively.

Now, `AB = ( beta - alpha ) hat i + (gamma - beta ) hat j +(alpha - gamma ) hatk`

`BC = (gamma - beta ) hat i + (alpha - gamma ) hat j + ( beta - alpha ) hat k`

` AC = (gamma - alpha ) hat i + (alpha - beta ) hat j + ( beta - gamma ) hat k`

`=> |AB| = sqrt( ( beta - alpha )^2 + (gamma - beta )^2 +(alpha - gamma )^2)`

` = sqrt( ( alpha - beta )^2 + (beta - gamma )^2 +( gamma - alpha )^2)` ......(1)

`|BC| = sqrt((gamma - beta )^2 + (alpha - gamma )^2 + ( beta - alpha )^2 )`

` = sqrt( ( alpha - beta )^2 + (alpha - gamma )^2 +( gamma - alpha )^2)` ......(2)

and `|AC| = (gamma - alpha )^2 + (alpha - beta )^2 + ( beta - gamma )^2 ) `

` = sqrt( ( alpha - beta )^2 + (beta - gamma )^2 +( gamma - alpha )^2)` ......(3)

From, Eqs. (i), (ii) and (iii), we get

`| AB | = | BC | = | AC |`

Hence, `ABC` form an equilateral triangle.
Correct Answer is `=>` (B) form an equilateral triangle
Q 1619591410

The line joining the points `(2, 1, 3)` and `(4, -2, 5)` cuts the
plane `2x + y - z = 3.`

What is the ratio in which the plane divides the line?
NDA Paper 1 2014
(A)

`1 : 1`

(B)

`2: 3`

(C)

`3 : 4`

(D)

None of these

Solution:

Let the ratio plane divides the line is `k : 1`.

Then, `0 = (4k+2)/(k+1)`

` => 4k + 2 = 0 => k = - 1/2`

and `4= (-2k + 1) /(k+1) => 4k + 4 = -2k + 1 => k = - 1/2`

Hence, plane divides the line in ratio `1 : 2` externally.
Correct Answer is `=>` (D) None of these
Q 2389234117

What is the distance of the point `( 1, 2, 0 )` from
`YZ`-plane is
NDA Paper 1 2013
(A)

1 unit

(B)

2 units

(C)

3 units

(D)

4 units

Solution:

Since, the point `(0, 2, 0)` lie on `YZ` -plane corresponding
to the point `(1, 2, 0)` .

`:. ` Distance between `(1, 2, 0)` and `(0, 2, 0)`

`= sqrt ( (1-0)^2 + (2-2)^2 + (0-2)^2) = sqrt (1+0 +0) =1 .`
Correct Answer is `=>` (A) 1 unit
Q 2387178087

If the distance between the points `(7, 1, -3)` and
`(4, 5, lambda )` is `13` units, then what is one of the values
of `lambda`?
NDA Paper 1 2012
(A)

`20`

(B)

`10`

(C)

`9`

(D)

`8`

Solution:

Given, the distance between the points

`(7, 1,- 3)` and `(4, 5, lambda) = 13`

`=> sqrt ((4-7)^2 + (5-1)^2 + (lambda+3)^2) =13`

`=> sqrt ( (-3)^2 + (-4)^2 + (lambda +3)^2) = 13`

`=> sqrt (9+ 16 + (lambda +3)^2) =13`

`=> sqrt (25+ (lambda +3)^2) =13`

On squaring both sides, we get

`25 + (lambda +3)^2 =169`

`=> 25 +lambda^2 +9 + 6 lambda -169 = 0`

`=> lambda^2 + 6 lambda -135 = 0`

`=> lambda^2 +15 lambda -9 lambda -135 =0`

`=> lambda (lambda + 15) -9 (lambda +15) = 0`

`=> (lambda +15) (lambda-9) =0 => lambda = 9 , -15`
Correct Answer is `=>` (C) `9`
Q 2309367218

If `O(0,0,0), P(3,4,5), Q(m,n,r)` and `R( 1, 1, 1)` be
the vertices of a parallelogram taken in order, then
what is the value of `m + n + r`?
NDA Paper 1 2010
(A)

`6`

(B)

`12`

(C)

`15`

(D)

More than `15`

Solution:

We know that in a parallelogram, diagonals bisect
each other.

Mid-point of `OQ =` Mid-point of `PR`

`=> ((0+m)/2 , (0+n)/2 , (0+r)/2) equiv ((1+3)/2 , (1+4)/2 , (1+5)/2)`

`=> m = 4, n = 5` and `r = 6`

`:. m + n - i - r = 4 + 5 + 6 = 15 `
Correct Answer is `=>` (C) `15`
Q 2349591413

What is the ratio in which the line joining the
points `(2, 4, 5)` and `(3, 5, -4)` is internally divided by
the `XY`-plane?
NDA Paper 1 2007
(A)

`5 :4`

(B)

`3:4`

(C)

`1:2`

(D)

`7:5`

Solution:

Let the line joining the points `(2, 4, 5)` and `(3, 5, - 4)` is
internally divided by the `XY`-plane in the ratio `m : n`, then its
`z`-coordinates should be zero.

`:. 0 = (-m xx 4 + 5n)/(m+n)` (by internal section formula)

`=> 4m=5n => m:n=5:4`
Correct Answer is `=>` (A) `5 :4`
Q 2349378213

Find the coordinate of the point, which is
equidistant from the points `(0, 0 0) , (2,0,0),
(0, 4, 0)`, and `(0, 0, 6)`, respectively
NDA Paper 1 2008
(A)

`(1, 2, 3)`

(B)

`(2, 3, 1)`

(C)

`(3, 1, 2)`

(D)

`(1,3,2)`

Solution:

Let the equation of sphere which passes through the
given four points be


`x^2 + y^2 + z^2 + 2ux + 2vy + 2wz +d =0`

`:. 0+0+0+0+0+0 + d =0`

`=> d =0` ..........(i)

So, `(2)^2 + 0 + 0 + 4u + 0+0 +d =0`

`4u + d = -4`

`=> 4u + 0 = -4 => u = -1` .......................(ii)

Therefore,

`0 + (4)^2 + 0+ 0 + 8v + 0 +d =0`

`=> 16 + 8v + d =0`

`=> 8v + 0 = -16 => v =-2` ...........(iii)

Now, `0+0 + (6)^2 + 0+ 0 + 12w +d =0`

`=> 36 +12w +d =0`

`=> 12w + 0 = -36 => w = -3`.................(iv)

`:.` Centre of sphere `= (1, 2, 3)`

So, the required point is the centre of sphere, which is equidistant
from given four points.
Correct Answer is `=>` (A) `(1, 2, 3)`
Q 2135091862

The point equidistant from the points `(a, 0, 0), (0, b, 0),
(0, 0, c)` and `(0, 0, 0)` is

(A)

`(a/3,b/3,c/3)`

(B)

`(a,b,c)`

(C)

`(a/2,b/2,c/2)`

(D)

None of these

Solution:

Let point is `(alpha , beta, gamma)`

`:. (alpha - a)^2 + beta^2 + gamma^2 = alpha^2 + (beta -b)^2 + gamma^2`

`= alpha^2 + beta^2 + (gamma -c)^2 = alpha^2 + beta^2 + gamma^2`

we get, `alpha = a/2 , beta =b/2` and `gamma =c/2`

`:.` Required point is `(a/2, b/2 ,c/2)`
Correct Answer is `=>` (C) `(a/2,b/2,c/2)`
Q 2836601572

The ratio in which the line joining (2, 4, 5),
(3, 5,- 4) is divided by the YZ-plane is

(A)

2:3

(B)

3:2

(C)

-2:3

(D)

4:-3

Solution:

Let the required ratio be ` lambda : 1`. Then,
the point is

`( ( 3 lambda +2)/( lambda +1 ) ,( 5 lambda +4)/( lambda +1) , ( -4 lambda +5 )/(lambda +1) )`

It lies on YZ-plane, so its x-coordinate
of x = 0

i.e., `(3 lambda +2 )/( lambda +1) = 0 => lambda = -2/3`

So, the ratio is 2 : 3 externally.
Correct Answer is `=>` (A) 2:3
Q 2136391272

The value (s) of `lambda`, for which the triangle with vertices

`(6, 10, 10), (1, 0,- 5)` and `(6,-10, lambda)` will be a right angled

triangle is/are

(This question may have multiple correct answers)

(A) `1`
(B) `70/3`
(C) `35`
(D) `0`
Solution:

Let the given points `A, B` and `C` respectively. Then

`AB^2 = 350, AC^2 = 500- 20 lambda + lambda^2 , BC^2 = 150 + 10 lambda + lambda^2 `

Now, `AB^2 + AC^2 = BC^2`

`=> 350 + 500- 20 lambda + lambda^2 = 150 + 10 lambda + lambda^2`

`=> lambda =70/3`

Next , `AB^2 + BC^2 = AC^2`

`=> 350 + 150 + 10 lambda + lambda^2 = 500- 20 lambda + lambda^2`

`=> lambda = 0`

Further, `BC^2 + AC^2 = AB^2`

`=> 150 + 10 lambda + lambda^2 + 500- 20 lambda + lambda^2 = 350`

`=> lambda^2 - 5 lambda + 150 = 0`,

Which have no real solution.


`:.` The triangle is right angles for `lambda = 0, 70/3`.
Correct Answer is `=>` (B)
Q 1967545485

Find the equation of the set of points `P`, which
are equidistant from `(1, 2, 3)` and `(3, 2,-1)`.
Class 11 Exercise 12.2 Q.No. 4
Solution:

Let `A( 1, 2, 3)` and `B(3, 2,- 1)` be the given points

and the coordinates of points `P` be `(x, y, z)`

We are given `AP = BP` or `AP^2 = BP^2`

`| (x-1)^2 + (y-2)^2 + (z - 3)^2`

`= (x-3)^2 + (y - 2)^2 + (z + 1 )^2 |`

or `x^2 + y^2 + z^2 - 2x - 4y - 6z + 1 + 4 + 9`

`= x^2 + y^2 + z^2 - 6x- 4y + 2z + 9 + 4 + 1`

or `-2x -4y -6z = -6x-4y + 2z`

or `4x - 8z = 0` or `x - 2z = 0`
Q 2359512414

What is the distance of the line `2x + y + 2z = 3`
from the origin?
NDA Paper 1 2013
(A)

`1` unit

(B)

`1.5` units

(C)

`2` units

(D)

`2.5` units

Solution:

Distance of the line `2x + y + 2z = 3` from the origin

` = ( |2 xx 0 + 0 +2 xx 0 -3 | )/( sqrt ( (2)^2 + (1)^2 + (2)^2) )`

`= ( |0+0+0 -3 | )/(sqrt (4+1+4) ) = (|-3|)/(sqrt(9))`

`= 3/3 =1` unit
Correct Answer is `=>` (A) `1` unit
Q 2440201113

If the sum of the squares of the distances of the
point `(x, y, z)` from the points `(a, 0, 0)` and `(-a, 0, 0)`
is `2c^2` , then which one of the following is correct?
NDA Paper 1 2007
(A)

`x^2 + a^2 = 2c^2 - y^2- z^2`

(B)

`x^2 + a^2 = c^2 - y^2 - z^2`

(C)

` x^2 - a^2 = c^2 - y^2 - z^2`

(D)

`x^2 + a^2 = c^2 + y^2 + z^2`

Solution:

Since, the sum of squares of the distances of the point
`P(x, y, z)` from the points `A(a, 0, 0)` and `B(-a, 0, 0)` is `2c^2` .

`:. PA^2 + PB^2 = 2c^2`

`=> (x - a)^2 + y^2 + z^2 + (x + a)^2 + y^2 + z^2 = 2c^2`

`=>x^2+ a^2 -2xa+ 2y^2 + 2z^2 + x^2 + a^2 + 2xa =2c^2`

`=> x^2 + a^2 + y^2 + z^2 = c^2`

`=> x^2 + a^2 = c^2 -y^2 - z^2`
Correct Answer is `=>` (B) `x^2 + a^2 = c^2 - y^2 - z^2`
Q 2369167015

What is the distance of the origin from the plane
`2x + 6 y - 3z + 7 = 0` ?
NDA Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`6`

Solution:

The distance of the point `P(x_1 , y_1)` from the plane

`ax + by + cz + d = 0` is ` | (ax_1 + by_1 +cz_1 +d)/(sqrt (a^2 +b^2 + c^2) ) |`

`:.` Required distance `= |7/(sqrt (2^2 +6^2 +3^2) ) |`

`= |7/(sqrt (4+36 +9) ) | = |7/7 | = 1`
Correct Answer is `=>` (A) `1`

Direction cosine and direction ratio

Q 2783191047

A straight line with direction cosines `(0, 1, 0)` is
NDA Paper 1 2017
(A)

parallel to x-axis

(B)

parallel to y-axis

(C)

parallel to z-axis

(D)

equally inclined to all the axes

Solution:

A straight line with direction cosines `(0, 1, 0)` only be parallel to y-axis due to law of symmetricity.
Correct Answer is `=>` (B) parallel to y-axis
Q 2231434322

The lines `2x = 3y = - z` and `6x = - y = - 4z`
NDA Paper 1 2015
(A)

are perpendicular

(B)

are parallel

(C)

intersect at an angle `45^0`

(D)

intersect at an angle `60^0`

Solution:

Here ` x/(1//2) = y/(1//3) = z/(-1)` and `

x/(1//6) = y/(-1) = z/(-1//4)`

DR's of the lines are `< 1/2 , 1/3 ,- 1 >` and

`< 1/6 ,-1 ,- 1/4 >`

i.e.`1/2 xx 1/6 + 1/3 (-1) + (-1) (- 1/4 ) =

1/(12) - 1/3 +1/4 = 0`

Hence, both lines are perpendicular.
Correct Answer is `=>` (A) are perpendicular
Q 1648280103

The projections of a directed
line segment on the coordinate axes are `12, 4, 3`,
respectively.

What are the direction cosines of the line segment?
NDA Paper 1 2015
(A)

`(pm (12)/(13) , pm (4)/(13) , pm (3)/(13) )`

(B)

`( (12)/(13) , - (4)/(13) , (3)/(13) )`

(C)

`( (12)/(13) , - (4)/(13) , - (3)/(13) )`

(D)

`(- (12)/(13) , - (4)/(13) , (3)/(13) )`

Solution:

Direction consine of the line segment

`(pm (12)/(13) , pm (4)/(13) , pm (3)/(13) )`

` { :. ( pm a)/sqrt( a^2 + b^2 + c^2 ) , ( pm b)/sqrt( a^2 + b^2 + c^2 ) , ( pm c)/sqrt( a^2 + b^2 + c^2 ) }`
Correct Answer is `=>` (A) `(pm (12)/(13) , pm (4)/(13) , pm (3)/(13) )`
Q 1668280105

From the point `P(3, -1, 11)`,
a perpendicular is drawn on the line `L` given by the
equation `x/2 = (y-2)/3 = (z-3)/4 `. Let `Q` be the foot of the
perpendicular.

What are the direction ratios of the line segment
`PQ?`
NDA Paper 1 2015
(A)

`(1, 6, 4)`

(B)

`( -1, 6, -4)`

(C)

`(-1, -6, 4)`

(D)

`(2, -6, 4)`

Solution:

We have,

`x/2 = (y-2)/3 = (z-3)/4 = lamda`

` = x=2lamda , y = 3 lamda + 2,z =4 lamda +3`

:. Direction ratios of the line passing through P

`P (3, -1, 11)` is `[(2 lamda - 3) (3 lamda + 2 + 1) (4 lamda + 3 - 11) ]`

`= 2 lamda - 3, 3 lamda + 3, 4 lamda - 8`

`:. (2 lamda - 3) . 2 + (3 lamda + 3) . 3 + (4 lamda - 8) . 4 = 0`

`=> 4 lamda - 6 + 9 lamda + 9 + 16 lamda - 32 = 0`

`=> 29 lamda = 29`

` => lamda = 1`

Direction ratios `= (2 - 3, 3 + 3 , 4 - 8)`

`=(-1,6,-4)`
Correct Answer is `=>` (B) `( -1, 6, -4)`
Q 2211734620

The direction ratios of the line perpendicular to the lines
with direction ratios `< 1, - 2, - 2 >` and `< 0, 2, 1 >` are
NDA Paper 1 2015
(A)

`< 2,-1,2 >`

(B)

`< -2,1,2 >`

(C)

`< 2, 1, - 2 >`

(D)

`< - 2,- 1,- 2 >`

Solution:

Let DR's of the line be `a, b, c`.

We have, `a- 2b -2c = 0` and `0 . a+ 2b + c = 0`

`:. a/(-2+4) = (-b)/(1-0) = c/(2-0) `

` => a/2 = b/(-1) = c/ 2`
Correct Answer is `=>` (A) `< 2,-1,2 >`
Q 2319567419

What is the angle between any two diagonals of
the cube'?
NDA Paper 1 2010
(A)

`cos^(-1) (1/2)`

(B)

`cos^(-1) (1/3)`

(C)

`cos^(-1) (1/sqrt(3))`

(D)

`cos^(-1) (2/sqrt(3))`

Solution:

Direction ratio of diagonal `OP`

`= 2 - 0, 2- 0, 2 - 0 = < 2, 2, 2 >`

and direction cosines `= (: 2/(2 sqrt(3)) , 2/(2 sqrt(3) ) , 2/(2 sqrt(3)) :) = (: 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) :)`

Direction ratio of diagonal `AB`

` = 2 - 0, 0- 2, 2 - 0 = < 2, -2,2 >`

and direction cosines `= (: 2/(2 sqrt(3)) , -2/(2 sqrt (3) ) , 2/(2 sqrt(3)) :)`

`= (: 1/sqrt(3) , -1/sqrt(3) ,1/sqrt(3) :)`

Let `theta` be the angle between `OP` and `AB`

`cos theta= (1/sqrt(3)) (1/sqrt(3)) + (1/sqrt(3)) (-1/sqrt(3)) + (1/sqrt(3)) (1/sqrt(3))`

`= 1/3 -1/3 +1/3`

`=> theta= cos^(-1) (1/3)`
Correct Answer is `=>` (B) `cos^(-1) (1/3)`
Q 1743756643

A straight line passes through
`(1, -2, 3)` and perpendicular to the plane `2x + 3y - z = 7`.

What are the direction ratios of normal to plane?
NDA Paper 1 2014
(A)

`< 2, 3, - 1 >`

(B)

`< 2, 3, 1 >`

(C)

`< -1, 2, 3 >`

(D)

None of these

Solution:

Given that, a line passes through the point `(1, -2, 3)` and

perpendicular to the plane `2x + 3y - z = 7` i.e., this straight

line is a normal to the plane.

Equation of the plane, `2x + 3y - z = 7`

On compaire with `ax + by + cz = d`,

`a = 2 , b = 3` and `c = -1`

which is the required direction ratios of normal to the plane

i.e.,` < a ,b , c > = < 2 ,3 ,-1 >`
Correct Answer is `=>` (A) `< 2, 3, - 1 >`
Q 2703291148

Under which one of the following conditions are the lines `x = ay + b; z=cy+d` and `x=ey+f; z=gy+h `perpendicular?
NDA Paper 1 2017
(A)

`ae+cg-1=0`

(B)

`ae+ bf -1 = 0`

(C)

`ae+cg+1=O`

(D)

`ag+ce+ 1 = 0`

Solution:

Given line

`(x-b)/a =(z-d)/c =(y-0)/1..............(l_1)`

`(x-f)/e =(x-h)/g =(y-0)/1 ..............(l_2)`

If `l_1 bot l_2` Then multiplication of their cosine will be zero

`ae +cg +1=0`
Correct Answer is `=>` (C) `ae+cg+1=O`
Q 1639591412

Consider the plane passing through the points `A(2, 2, 1),
8(3, 4, 2)` and `C(7, 0, 6)`.

What are the direction ratios of the normal to the plane?
NDA Paper 1 2014
(A)

`< 1,0, 1 >`

(B)

` < 0, 1,0 >`

(C)

`< 1, 0, -1 >`

(D)

None of these

Solution:

We know that, equation of plane passing

through three non-collinear points `(x_1, y_1, z_1), (x_2 , y_2, z_2)`

and `(x_3 , y_3 , z_3 )` is

` |(x - x_1, y - y_1, z - z_1),(x_2 - x_1 , y_ 2 - y_1, z_2 - z_1),( x_3 - x_1, y_3 - y_1, z_3 - z_1)| = 0`

Put the value of `(x_1, y_1, z_1), (x_2 , y_2, z_2 )` and `(x_3 , y_3, z_3 )`, we get

` |(x-2, y-2, z-1),(1, 2 ,1) ,(5, -2, 5)| = 0`

`=> (x -2)(10 + 2) - (y -2)(5- 5) + (z -1) (-2 -10) = 0`

`=> 12x - 12z =12 => x - z = 1`

Hence, the equation of plane passes through `(1, 0, 1 )`.

Direction ratios of the normal to the plane

`x - z = 1` are `(1, 0, -1)` .
Correct Answer is `=>` (C) `< 1, 0, -1 >`
Q 2572291136

The cosine of the angle between any two diagonals of a cube is
WBJEE 2016
(A)

`1/3`

(B)

`1/2`

(C)

`2/3`

(D)

`1/sqrt3`

Solution:

Let the direction ratio's of two diagonals of a cube are

`(1, 1, 1)` and `(-1, 1, 1)`.

` :. cos theta = (a_1a_2 + b_1b_2 + c_1c_2)/( sqrt(a_1^2 + b_1^2 + c_1^2) sqrt(a_2^2 + b_2^2 + c_2^2))`

` = ( (1) xx (-1) + (1) xx (1) + (1) xx (1) )/(sqrt( (1)^2 + (1)^2+ (1)^2 ) sqrt( (- 1)^2 + (1)^2 + ( 1)^2)`

` = ( -1 + 1 + 1)/(sqrt3 sqrt3) = 1/3`
Correct Answer is `=>` (A) `1/3`
Q 2344178053

The direction cosines of a line equally
inclined with co-ordinate axes are :
BITSAT Mock
(A)

`< 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) >`

(B)

`< 0, 1, 0 >`

(C)

`< 1, 0, 0, >`

(D)

`< 1, 1, 1 >`

Solution:

Let `α, β, γ` be angles made by line on

co-ordinate axes.

Since line is equally inclined to

co-ordinate axes.

`∴ α = β = γ`

`⇒ cos α = cos β = cos γ` ...(1)

Also `cos^2 α + cos^2 β + cos^2 γ = 1` ...(2)

from (1) and (2)

`cos^2 α + cos^2 α + cos^2 α = 1`

`⇒ 3 cos^2 α = 1`

`⇒ cos^2 α = 1/3`

`⇒ cos α = 1/sqrt(3)`

Since `cos α = cos β = cos γ = 1/sqrt(3)`

The direction cosines are `(1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3))`
Correct Answer is `=>` (A) `< 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) >`
Q 1713267149

Let a vector r make angles `60^0`,
`30^0` with `X` and `Y`-axes, respectively.

What angle does `r` make the `Z`-axis?
NDA Paper 1 2014
(A)

`30^0`

(B)

`60^0`

(C)

`90^0`

(D)

`120^0`

Solution:

Given that,

A vector `r` make an angle `60^0` with `X`-axis i.e., `l = cos 60^0 = 1/2`

and `a` vector `r` make an angle `30^0` with `Y`-axis

i.e.,` m = cos 30^0 = sqrt(3)/2`

Let the vector `r` make an angle `theta` with `Z`-axis i.e.,

`n = cos theta`

We know that,

`l_2 + m_2 + n_2 = 1`

`=> (1/2)^2 + (sqrt(3)/2) + cos^2 theta =1`

` (1/4 + 3/4 ) + cos^2 theta =1`

` => cos^2 theta = 1 - 1 = 0`

` => cos theta = 0 = cos 90^0`

` :. theta = 90^0`

Hence, ` 90^0` angle does `r` make with `Z`-axis.
Correct Answer is `=>` (C) `90^0`
Q 2309434318

What are the direction cosines of a line which is
equally inclined to the positive directions of the
axes?
NDA Paper 1 2012
(A)

` (: 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) :)`

(B)

` (: - 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) :)`

(C)

` (: - 1/sqrt(3) , - 1/sqrt(3) , 1/sqrt(3) :)`

(D)

` (: 1/3 , 1/3 , 1/3 :)`

Solution:

If `alpha , beta` and `gamma` are the angles that a line makes with the
coordinate axes.

Then, `l= cos alpha , m = cos beta` and `n =cos gamma`

`:. l^2 + m^2 + n^2 = 1`

`=> cos^2 alpha + cos^2 beta + cos^2 gamma =1` ... (i)

Here, `alpha = beta = gamma` (given)


From Eq. (i),

`cos^2 alpha + cos^2 alpha + cos^2 alpha = 1`

`=> cos alpha = 1/sqrt(3)`


(since, direction cosines of a line which is equally inclined to the
positive directions of the axis, so we take only positive sign)

`:. cos alpha = cos beta = cos gamma = 1/sqrt(3)`

or `l =m = n = 1/sqrt(3)`



Hence, the required direction cosines are ` (: 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) :)`
Correct Answer is `=>` (A) ` (: 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) :)`
Q 2359645514

What is the sum of the squares of direction
cosines of the line joining the points `(1, 2,-3)` and `(-2, 3, 1)`?
NDA Paper 1 2012
(A)

`0`

(B)

`1`

(C)

`3`

(D)

`2/(sqrt (26))`

Solution:

The equation of line passing through `(x_1 , y_1, z_1 )` and `(x_1, y_2 ,z_2)` is

`(x-x_1)/(x_2 -x_1) = (y-y_1)/(y_2 -y_1) = (z-z_1)/(z_2 - z_1)` ............(i)

Here, `(x_2 - x_1 ), (y_2 - y_1)` and `(z_2 - z_1)` direction ratio's of that
line.
Then, its direction cosines are

`l = (x_2 -x_1)/(sqrt ( sum (x_2 -x_1)^2 ) ) , m = (y_2 -y_1)/(sqrt (sum (x_2 -x_1)^2) )` and `n = (z_2 -z_1)/(sqrt (sum (x_2 -x_1)^2 ) ) n`

Here, `(x_1, y_1, z_1) = (1, 2,- 3)` and `(x_ 2, y_2, z_2) = (-2, 3, 1)`

`:. l = (-2-1)/(sqrt ( (-3)^2 + (1)^2 + (4)^2) )`

`m = (3-2)/(sqrt ( (-3)^2 + (1)^2 + (4)^2 ) )`

and `n= (1+3)/(sqrt ((-3)^2 + (1)^2 + (4)^2 ) )`

`=> l = -3/sqrt(26) , m =1/sqrt(26) , n = 4/sqrt(26)`

`:. l^2 +m^2 +n^2 = (-3/sqrt(26))^2 + (1/sqrt(26))^2 + (4/sqrt(26))^2`

`= 9/26 + 1/26 + 16/26`

`= 26/26 =1`
Correct Answer is `=>` (B) `1`
Q 2369691515

A line makes angles `theta, phi` and `psi` with `X, Y` and
`Z`-axes, respectively.

Consider the following statements

I. `sin^2 theta + sin^2 phi = cos^2 psi`

II. `cos^2 theta + cos^2 phi = sin^2 psi`

III. `sin^2 theta + cos^2 phi = cos^2 psi`

Which of the above statements is/arc correct?
NDA Paper 1 2007
(A)

Only I

(B)

Only II

(C)

Only Ill

(D)

Both II and1111

Solution:

We know that, if a line makes angle `theta, phi` and `psi` with `X, Y`
and `Z`-axes, respectively.

`:. cos^2 theta + cos^2 phi + cos^2 psi =1`

`=> cos^2 theta + cos^2 phi =1- cos^2 psi`

`=> cos^2 theta + cos^2 phi = sin^2 psi`

Hence, only Statement II is correct.
Correct Answer is `=>` (B) Only II
Q 2329791611

What are the direction cosines lof the line
represented by

`3x + y + 2z = 7` and `x + 2 y + 3z = 5` ?
NDA Paper 1 2007
(A)

`(-1, -7, 5)`

(B)

`(-1, 7, 5)`

(C)

`(-1/sqrt(75) , - 7/(sqrt(75) ) , 5/(sqrt(75)))`

(D)

`(-1/sqrt(75) , 7/(sqrt(75) ) , 5/(sqrt(75)))`

Solution:

Let the direction ratio of the line which is passing
through the planes `3x + y + 2z = 7` and `x + 2Y + 3z = 5` be `a,
b, c`, then the condition between the direction ratio of this line and
the direction ratio of normal to the plane is

`3a + b + 2c = 0`

`a+ 2b + 3c = 0`

By cross-multiplication method,

`a/(-1) = b/(-7) =c/5`

`:.` The direction cosines of this line is

` (: (-1)/(sqrt (1+25+49) ) , (-7)/(sqrt(1+25+49)) , 5/(sqrt(1+25 +49)) :)`

`= (: (-1)/(sqrt(75)), (-7)/(sqrt(75)) , (5)/(sqrt(75)) :)`
Correct Answer is `=>` (C) `(-1/sqrt(75) , - 7/(sqrt(75) ) , 5/(sqrt(75)))`
Q 2309178018

What are the direction ratios of the line
determined by the planes `x - y + 2z = 1` and
`x+y-z=3` ?

NDA Paper 1 2009
(A)

`(-1, 3,2)`

(B)

`(-1,-3,2)`

(C)

`(2, 1, 3)`

(D)

`(2, 3,2)`

Solution:

The intersection of the given plane is

`x - y + 2z- 1 + lambda ( x + y- z- 3) = 0`

`=> x (1 + lambda ) + y(lambda - 1) + z(2 - lambda ) - 3 lambda - 1 = 0`

Direction ratios of normal to the above plane is

`(1+lambda , lambda -1 , 2- lambda)`

Since, the line formed intersected by planes and the normal of
the plane are perpendicular, then by taking option (a),

`-1 (1+lambda) + 3 (lambda -1) + 2 (2- lambda) =0`

`=> -1 - lambda +3 lambda -3 +4 -2 lambda =0`

`=> 0=0`
Correct Answer is `=>` (A) `(-1, 3,2)`
Q 2349178013

A line makes the same angle a with each of the `X`
and `Y`-axes. If the angle `theta`, which it makes with the
`Z`-axis, is such that `sin^2 theta = 2 sin^2 alpha`, then what is
the value of `alpha`?
NDA Paper 1 2009
(A)

`pi/4`

(B)

`pi/6`

(C)

`pi/3`

(D)

`pi/2`

Solution:

`:. l^2 + m^2 + n^2 =1`

i.e., `cos^2 alpha + cos^2 beta + cos^2 gamma = 1`

`=> cos^2 alpha + cos^2 alpha + cos^2 theta = 1 [ :. (alpha = beta ) , ( gamma = theta ) ]` - .(i)

Also, `sin^2 theta = 2 sin^2 alpha` (given)


`=> 1- cos62 theta = 2 (1- cos^2 alpha)`

`=> cos^2 theta = 2 cos^2 alpha -1`


From Eq. (i),

`2 cos^2 alpha + (2 cos^2 alpha -1) =1`

`=> 4 cos^2 alpha =2 => cos^2 alpha =1/2`

`=> cos alpha = pm 1/sqrt(2) => alpha = pi/4 , (3 pi)/4`
Correct Answer is `=>` (A) `pi/4`
Q 2349067813

What is the angle between the diagonal of one of
the faces of the cube and the diagonal of the cube
intersecting the diagonal of the face of the cube?
NDA Paper 1 2010
(A)

`cos^(-1) (1/sqrt(3))`

(B)

`cos^(-1) (2/sqrt(3))`

(C)

`cos^(-1) (sqrt(2/3))`

(D)

`cos^(-1) (sqrt(2)/3)`

Solution:

Direction ratio of side `OC = 2 - 0, 2 - 0, 0 - 0`

`= (: 2,2,0 :)`

and direction cosines `= (: 2/(2 sqrt(2) , 2/(2 sqrt(2)) , 0) :) , (: 1/sqrt(2) , 1/sqrt(2) , 0 )`

Let `theta_2` be the angle between side `OC` and diagonal `OP`

`cos theta_2 = (1/sqrt(2)) (1/sqrt(3)) + (1/sqrt(2)) (1/sqrt(3)) + (0) (1/sqrt(3))`

`= 1/sqrt(6) + 1/sqrt(6) + 0 = 2/sqrt(6) = sqrt(2/3)`

`=> theta_2 = cos^(-1) (sqrt(2/3))`
Correct Answer is `=>` (C) `cos^(-1) (sqrt(2/3))`
Q 2309067818

If the line through the points `A (k, 1, -1)` and
`B (2k, 0, 2)` is perpendicular to the line through the
points `B` and `C(2 + 2k, k, 1)`, then what is the value
of `k`?
NDA Paper 1 2010
(A)

`-1`

(B)

`1`

(C)

`-3`

(D)

`3`

Solution:

Direction ratio of line `AB`

`= 2k - k, 0 - 1, 2 + 1 = (: k, - 1, 3 :)`

and direction ratio of line `BC`

`= 2 + 2k - 2k, k - 0, 1 - 2`

`= (: 2, k, -1 :)`

Since, both the lines are perpendicular.

`:. (2)(k) + (-1)(k) + (3)(-1) = 0`

`=> 2k - k - 3 = 0 => k=3`
Correct Answer is `=>` (D) `3`
Q 2319156910

Consider the following statements among the
angles `alpha , beta` and `gamma` made by a veftor with the
coordinate axes I


I. `cos 2alpha + cos 2 beta + cos 2 gamma = -1`

II. `sin^2 alpha + sin^2 beta + sin^2 gamma = 1`

Which of the abmJe statements is/are correct?
NDA Paper 1 2011
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I ror II

Solution:

`:. cos^2 alpha + cos^2 beta + cos^2 gamma = 1`

`=> 2 cos^2 alpha + 2 cos^2 beta + 2 cos^ 2 gamma = 2`


`=> 2 cos^2 alpha - 1 + 2 cos^2 beta - 1 + 2 cos^2 gamma - 1 = 2 - 3`

`=> cos 2 alpha + cos 2 beta + cos 2 gamma = - 1`

Now,

`1 - sin^2 alpha + 1 - sin^2 beta + 1 - sin^2 gamma = 1`

`=> sin^2 alpha + sin^2 beta + sin^2 gamma = 2`

Hence, only Statement I is correct.
Correct Answer is `=>` (A) Only I
Q 2389745617

If a line makes the angles `alpha , beta ` and `gamma` with the axes,
then what is the value of

`1 + cos 2 alpha + cos 2 beta + cos 2 gamma ?`
NDA Paper 1 2012
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Given that `alpha , beta` and `gamma` are angles, which makes a line
with the axes, then

`cos^2 alpha + cos^2 beta + cos^2 gamma =1`

`=> 2 cos^2 alpha + 2 cos^2 beta + 2 cos^2 gamma =2`

`=> (1+ cos 2 alpha) + (1+ cos 2 beta) + (1+ cos 2 gamma) = 2`

`=> 1+ cos 2 alpha + cos 2 beta + cos 2 gamma = 2-2 =0`
Correct Answer is `=>` (B) `0`
Q 2339845712

What are the direction ratios of normal to the
plane `2 - y + 2z + l = 0?`
NDA Paper 1 2012
(A)

`(: 2,1,2 :)`

(B)

` (: 1, -1/2 ,1 :)`

(C)

`(: 1,-2,1 :)`

(D)

None of these

Solution:

Equation of plane,

`2x-y+2z+1=0`

`=> x -1/2 y + z +1/2 =0`

`:.` Direction ratio's of normal to the plane `= < 1 , -1/2 ,1 >`
Correct Answer is `=>` (B) ` (: 1, -1/2 ,1 :)`
Q 2806823778

Consider the following relations among the
angles a., [3 and y made by a vector with the
coordinate axes.

I. `cos 2 alpha + cos 2 beta + cos 2 gamma = - 1`

II. `sin^2 alpha +sin^2 beta +sin^2 gamma = 1`

Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We know that,

`cos^2 alpha + cos^2 beta + cos^2 gamma =1` ...........(i)

I. `cos 2 alpha + cos 2 beta + cso 2 gamma `

`= 2 cos^2 alpha + 2 cos^2 beta + 2 cos^2 gamma -3`

`= 2 (cos^2 alpha + cos^2 beta + cos^2 gamma) -3 = 2 (1) -3 = -1`

So, Statement is true. [from Eq. (i)]

II. `sin^2 alpha + sin^2 beta + sin^2 gamma`

`= 1- cos^2 alpha + 1- cos^2 beta +1- cos^2 gamma`

`= 3 - (cos^2 alpha + cos ^2 beta + cos^2 gamma) = 3-1 = 2` [from Eq. (i)]

So, Statement II is false.
Correct Answer is `=>` (A) Only I
Q 2339223112

The sum of the direction cosines of `Z`-axis is
NDA Paper 1 2013
(A)

`0`

(B)

`1/3`

(C)

`1`

(D)

`3`

Solution:

We know that, the direction cosines of `Z`-axis are

`(0, 0, 1)`.

So, sum of the direction cosines of `Z` -axis `= 0 + 0 + 1 = 1`
Correct Answer is `=>` (C) `1`
Q 2349023813

If a line makes `30^(circ)` with the positive direction of
`X`-axis, `angle beta` with the positive direction of `Y`-axis and
` angle gamma` with the positive direction of `Z`- axis , then what
is `cos^2 beta + cos^2 gamma` equal to?
NDA Paper 1 2013
(A)

`1/4`

(B)

`1/2`

(C)

`3/4`

(D)

`1`

Solution:

We know that, if a line makes an `angle alpha` with the positive
direction of `X`-axis, `angle beta` with the positive direction of `Y`axis and `angle gamma`
with the positive direction of `Z`-axis, then

`cos^2 alpha + cos^2 beta + cos^2 gamma =1`

Given that, `alpha =30^(circ)`

`:. cos^2 30^(circ) + cos^2 beta + cos^2 gamma =1`

`=> (sqrt(3)/2)^2 + cos^2 beta + cos^2 gamma =1`

`=> cos^2 beta + cos^2 gamma = 1 -3/4`

`:. cos^2 beta + cos^2 gamma = 1/4`
Correct Answer is `=>` (A) `1/4`
Q 2449780613

The angle between the lines whose direction cosines are `(sqrt3/4 , 1/4 , sqrt3/2)`and `(sqrt3/4 , 1/4 , - sqrt3/2)` is
BCECE Stage 1 2012
(A)

`pi`

(B)

`pi/2`

(C)

`pi/3`

(D)

`pi/4`

Solution:

`costheta = |l_1l_2+m_1m_2+n_1n_2|`


` = |sqrt3/4xxsqrt3/4+1/4xx1/4+sqrt3/2xx(-sqrt3/2)|`


` = |3/16+1/16-3/4| = |-2/4| = 1/2`


`=> theta = pi/3`
Correct Answer is `=>` (C) `pi/3`
Q 2319312219

What is the sum of the squares of direction
cosines of `X`-axis?
NDA Paper 1 2013
(A)

`0`

(B)

`1/3`

(C)

`1`

(D)

`3`

Solution:

We know that, the direction cosines of `X`-axis is

`<1 0, 0 >`.

`:.` Sum of squares of direction cosine `= (1)^2 + (0)^2+ (0)^2`


`= 1 + 0 + 0 = 1`
Correct Answer is `=>` (C) `1`
Q 1723367241

Let a vector r make angles `60^0`,
`30^0` with `X` and `Y`-axes, respectively.

What are the direction cosines `r`?
NDA Paper 1 2014
(A)

`< 1/2 , sqrt(3)/2 , 0 >`

(B)

`< 1/2 , - sqrt(3)/2 , 0 >`

(C)

`< 1/sqrt(2) , 1/sqrt(2) , 0 >`

(D)

`< -1/2 , sqrt(3)/2 , 0 >`

Solution:

Given that,

A vector `r` make an angle `60^0` with `X`-axis i.e., `l = cos 60^0 = 1/2`

and `a` vector `r` make an angle `30^0` with `Y`-axis

i.e.,` m = cos 30^0 = sqrt(3)/2`

The direction cosine of the vector

` r = < l , m, n >`

`=> r = < cos 60^0 , cos 30^0, cos 90^0 >`

` => r = < 1/2 , sqrt(3)/2 , 0 >`
Correct Answer is `=>` (A) `< 1/2 , sqrt(3)/2 , 0 >`
Q 1753167944

If a line passes through the points `(6,- 7, -1)` and
`(2, - 3, 1)`, then what are the direction ratios of the
line?
NDA Paper 1 2014
(A)

`< 4, - 4, 2 >`

(B)

`< 4, 4, 2 >`

(C)

`< -4, 4, 2 >`

(D)

`< 2, 1, 1 >`

Solution:

If a line passes through the points

`(x_1, y_1, z_1)` and `(x_2 , y_2 , z_2 )`, then its direction ratio is

`< x_2 - x_1 , y_2 - y_1, z_2 - z_1 >`.

:. The direction ratio of the line which passes through the

points `(6, - 7, - 1)` and `(2, - 3, 1)` is

`< 2 - 6 , - 3 + 7,1 +1 > = < -4 , 2 , 2 >`
Correct Answer is `=>` (C) `< -4, 4, 2 >`
Q 1864780655

If a line has the direction ratios `- 18, 12, -4`
then what are its direction cosines?
Class 12 Exercise 11.1 Q.No. 3
Solution:

Now given direction ratios of a line are

`-18, 12,-4`

`:. a = -18, b = 12, c = -4`

Direction cosines are

`l = a/sqrt( a^2 + b^2 + c^2) = (-9)/(11)`

` m = b/sqrt( a^2 + b^2 + c^2) = 6/(11) `

` n = c/sqrt( a^2 + b^2 + c^2) = (-2)/(11) `

Hence direction cosines ` (-9)/(11) , 6/(11) , (-2)/(11) `
Q 2846812773

Direction cm;ines of the line which is
perpendicular to the lines whose direction ratios
are 1.- 1, 2 and 2, 1,- 1, are given by

(A)

`[ -1/(sqrt 35) ,5/(sqrt 35), 3/(sqrt 35 ) ]`

(B)

`[ -1/(sqrt 35) , -5/(sqrt 35), 3/(sqrt 35 ) ]`

(C)

`[ 1/(sqrt 35) ,5/(sqrt 35), - 3/(sqrt 35 ) ]`

(D)

None of these

Solution:

Let (a, b, c) be the direction ratios of given line.

`:. a -b + 2c = 0` ................(i)

`2a + b - c = 0` ............(ii)

From Eqs. (i) and (ii), we get

`a/(-1) = b/5 = c/3`

So, (-1, S, 3) arc the direction ratios of given line.
Hence, direction cosines will be

`((-1)/(sqrt 35) , 5/(sqrt 35) ,3/(sqrt 35) )`
Correct Answer is `=>` (A) `[ -1/(sqrt 35) ,5/(sqrt 35), 3/(sqrt 35 ) ]`
Q 2344178053

The direction cosines of a line equally
inclined with co-ordinate axes are :
BITSAT Mock
(A)

`< 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) >`

(B)

`< 0, 1, 0 >`

(C)

`< 1, 0, 0, >`

(D)

`< 1, 1, 1 >`

Solution:

Let `α, β, γ` be angles made by line on

co-ordinate axes.

Since line is equally inclined to

co-ordinate axes.

`∴ α = β = γ`

`⇒ cos α = cos β = cos γ` ...(1)

Also `cos^2 α + cos^2 β + cos^2 γ = 1` ...(2)

from (1) and (2)

`cos^2 α + cos^2 α + cos^2 α = 1`

`⇒ 3 cos^2 α = 1`

`⇒ cos^2 α = 1/3`

`⇒ cos α = 1/sqrt(3)`

Since `cos α = cos β = cos γ = 1/sqrt(3)`

The direction cosines are `(1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3))`
Correct Answer is `=>` (A) `< 1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3) >`

Equation of line and angle between two line

Q 1618280109

A triangular plane `ABC`
with centroid `(1, 2, 3)` cuts the coordinate axes at `A, B, C`,
respectively.

What are the intercepts made by the plane `ABC`
on the axes?
NDA Paper 1 2015
(A)

`3, 6, 9`

(B)

`1, 2, 3`

(C)

`1, 4, 9`

(D)

`2, 4, 6`

Solution:

We have, ` x/A + y/B + c/Z =1`

Since, triangular plane `ABC` with centroid `(1, 2, 3)` cuts

the coordinate axes, :. Intercepts made by

`:. X` axis, `Y`-axis and `Z`-axis are, `3, 6, 9`.
Correct Answer is `=>` (A) `3, 6, 9`
Q 2271734626

What are the coordinates of the foot of the perpendicular drawn from the point `(3, 5, 4)` on the plane `z = 0?`
NDA Paper 1 2015
(A)

`(0, 5, 4)`

(B)

`(3,5, 0)`

(C)

`(3,0, 4)`

(D)

`(0, 0, 4)`

Solution:

The foot of perpendicular drawn from `(3, 5, 4)` on the

`z =0` is

` (x-3)/0 = (y-5)/0 = (z-4)/1 = - 4/1`

`=> x=3 ,y= 5 , z= 0`

Hence, `(x, y, z) = (3, 5, 0)`
Correct Answer is `=>` (B) `(3,5, 0)`
Q 1713156940

The vertices of a `Delta ABC` are `A(2, 3, 1)`,
`B( -2, 2, 0)` and `C(0, 1, -1)`.

What is the cosine of angle `ABC?`
NDA Paper 1 2014
(A)

`1 /sqrt(3)`

(B)

`1/sqrt(2)`

(C)

`2/sqrt(6)`

(D)

None of these

Solution:

Given that vertices of a triangle are,

let `(x_1 , y_1 , z_1) = A (2, 3, 1)`,

`(x_2 ,y_2 , z_2 ) = B (- 2, 2, 0)`,

and `(x_3, y_3, z_3) = C (0, 1, -1)`,

Now, of DR's `AB = < a_1 , b_1 ,c_1 >`

` = < - 2 - 2, 2 - 3, 0 - 1 >`

`= < - 4, - 1, - 1 >`

and DR's of `BC = < a_2 , b_2 , c_2 >`

` = < 0 + 2, 1 - 2, - 1 - 0 >`

` = < 2 ,- 1, -1 >`

Let e be the angle between `AB` and `BC`.

`:. cos theta = | ((a_1a_2 +b_1b_2 +c_1 c_2 ) )/sqrt(((a_1^2 + b_1^2 + c_1^2)) sqrt( a_2^2 + b_2^2 + c_2^2)) |`

` = | { ( - 4 xx 2+(-1) (- 1) + (-1) xx (-1) )/sqrt( ((16 + 1 + 1)) sqrt(4 + 1+ 1)) }|`

` = | (-8 +1 +1 )/(sqrt(16) sqrt(6) )|`

` = | (-6)/(6sqrt(3)) | = | (-1)/sqrt(3) | = 1/sqrt(3)`

` :. cos theta =cos angle ABC = 1/ sqrt(3)`
Correct Answer is `=>` (A) `1 /sqrt(3)`
Q 2309112018

If `theta` is the acute angle between the diagonals of a
cube, then which one of the following is correct?
NDA Paper 1 2013
(A)

`theta = 30^(circ)`

(B)

`theta = 45^(circ)`

(C)

`2 cos theta =1`

(D)

`3 cos theta =1`

Solution:

Let `a` be length of the edge of the cube.

Let direction ratio's of the diagonal `OP` of the cube is `< a_1 b_1 c_1 >`

`= < a - 0, a- 0, a- 0>`

`= < a, a, a >`

and direction ratio's of the diagonal `AD` of the cube is

`< a_2, b_2, c_2 >`

`= < 0 - a, a - 0, a - 0 >`

`= < - a , a, a >`

Let e is the acute angle between the diagonals `OP` and `AD` of a
cube

`:. cos theta = | (a_1a_2 +b_1b_2 +c_1c_2)/(sqrt (a_1^2+b_1^2 + c_1^2) sqrt (a_2^2 +b_2^2 +c_^2)) |`

`= | ((a) (-a) +(a) (a) + (a) (a) )/(sqrt (a^2 + a^2 +a^2) sqrt(a^2 + a^2 +a^2) ) |`

`= | (-a^2 + a^2 + a^2 )/(sqrt (3a^2) sqrt (3a^2) ) | = |a^2/(3a^2) |`

`= 1/3 => 3 cos theta =1`
Correct Answer is `=>` (D) `3 cos theta =1`
Q 2319634510

What is the angle between the lines

`(x-2)/1 = (y+1)/-2= (z+2)/1` and `(x-1)/1 = (2y +3)/3 = (z+5)/2 ?`
NDA Paper 1 2012
(A)

`pi/2`

(B)

`pi/3`

(C)

`pi/6`

(D)

None of these

Solution:

Given lines are,

`(x-2)/1 = (y+1)/-2 = (z+2)/1`................(i)

and `(x-1)/1 =(2y +3)/3 = (z+5)/2`



`=> (x-1)/1 = (2 (y +3/2) )/3 = (z+5)/2`

`=> (x-1)/1 = (y +3/2)/(3/2) = (z+5)/2` .............(ii)

If `theta` be the acute angle between lines (i) and (ii). then

`cos theta = | (1 xx 1 + (-2) (3/2) + 1(2) )/(sqrt (1+ (-2)^2 +1^2) sqrt (1^2 + (3/2)^2 + 2^2 )) |`

`= | (1+ (-3) +2)/(sqrt (1+4+1) sqrt (1+ 9/4 +4) ) | = | 0/(sqrt(6) sqrt (29/4) ) | = 0`

`:. cos theta = 0 => theta = cos^(-1) (0) = pi/2``
Correct Answer is `=>` (A) `pi/2`
Q 2389134917

What is the equation to the straight line passing
through `(a, b, c)` and parallel to `Z`-axis?
NDA Paper 1 2012
(A)

`(x-a)/1 = (y-b)/0 = (z-c)/0`

(B)

`(x-a)/0 = (y-b)/0 = (z-c)/1`

(C)

`(x-a)/0 = (y-b)/1 = (z-c)/0`

(D)

`(x-a)/0 = (y-b)/1 = (z-c)/1`

Solution:

The equation to the straight line passing through
`(a, b, c)` and parallel to `Z`-axis is

`(x-a)/0 = (y-b)/0 = (z-c)/1`
Correct Answer is `=>` (B) `(x-a)/0 = (y-b)/0 = (z-c)/1`
Q 2359378214

If the angle between the lines with direction ratios

`(1,0 , pm cos alpha)` is `60^(circ)` , then what is the value of `alpha`?
NDA Paper 1 2008
(A)

`cos^(-1) (1/sqrt(2) )`

(B)

`cos^(-1) (1/sqrt(3) )`

(C)

`cos^(-1) (1/3 )`

(D)

`cos^(-1) (1/2 )`

Solution:

Angle between two lines whose direction ratios are

`a_1, b_1, c_1` and `a_2, b_2, c_2` respectively is `theta`.

Then , `cos theta = | (a_1a_2 + b_1b_2 + c_1c_2)/(sqrt(a_1^2 + b_1^2 + c_1^2) sqrt (a_2^2 + b_2^2 + c_2^2) ) |`

`:. cos 60^(circ) = | (1 xx 1 + 0 xx 0 + (cos alpha) (-cos alpha))/ (sqrt(1^2 + (0)^2 + cos^2 alpha) sqrt(1^2 + (0)^2 + (-cso alpha)^2 )) |`

`=> 1/2 = (1- cos^2 alpha)/(sqrt(1+cos^2 alpha ) sqrt (1+cos^2 alpha)) => 1/2 = (1- cos^2 alpha)/(1+ cos^2 alpha)`

`=> (1+2)/(1-2) = 2/(-2 cos^2 alpha)` (using componendo -dividedndo formula)

`=> 3/-1 = 1/(- cos^2 alpha)`

`=> cos alpha = 1/sqrt(3) => alpha = cos^(-1) (1/sqrt(3))`
Correct Answer is `=>` (B) `cos^(-1) (1/sqrt(3) )`
Q 2319156919

What is the angle between the lines whose
direction cosines are proportional to `(2, 3, 4)` and
`(1, -2, 1)`, respectively?
NDA Paper 1 2011
(A)

`90^(circ)`

(B)

`60^(circ)`

(C)

`45^(circ)`

(D)

`30^(circ)`

Solution:

` :. 2 xx 1 + 3 xx (-2) + 4 xx 1 = 0`

` ( :. cos theta = l_1 l_2 + m_1 m_2 + n_1 n_2 )`

`=> cos theta = 0 = cos 90^(circ) => theta = 90^(circ)`

So, angle between the lines is `90^(circ)`
Correct Answer is `=>` (A) `90^(circ)`
Q 2513580449

Cosine of the angle between two diagonals of a cube is equal to :
UPSEE 2013
(A)

`2/sqrt6`

(B)

`1/3`

(C)

`1/2`

(D)

None of these

Solution:

Let `OA, OB, OC` be the sides of a cube such that `OA =OB =OC =a`


`therefore` Co-ordinate of the vertices of cube are

`O(0, 0, 0), A(a, 0, 0), B(0, a, 0),`

`C(O, 0, a), D(a, a, 0), E(a, a, a), F(0, a, a)` and `G(a, 0, a)` .
`therefore` Direction ratios of `OE` are `(a- 0, a- 0, a- 0) i.e., (a, a, a)`.
`therefore` Direction cosines of `OE` are


`( a/sqrt(a^2+a^2+a^2) , a/sqrt(a^2+a^2+a^2) , a/sqrt(a^2+a^2+a^2))`


` = (1/sqrt3 , 1/sqrt3 , 1/sqrt3)`


Similarly, direction cosines of AF are

`therefore` Angle between OE and AF is


`cos^(-1) [-1/sqrt3 * 1/sqrt3 + 1/sqrt3 * 1/sqrt3 + 1/sqrt3*1/sqrt3]`

` = cos^(-1) (1/3)`
Correct Answer is `=>` (B) `1/3`
Q 2361480325

If the lines `(x-1)/2 = (y+1)/3 = (z-1)/4` and

`(x-3)/1 = (y-k)/2 = z/1` are intersecting

each other, then `k` is
BITSAT Mock
(A)

`2/9`

(B)

`9/2`

(C)

`1`

(D)

`3/2`

Solution:

Let `(x-1)/2 = (y+1)/3 = (z-1)/4 = lambda`

and `(x-3)/1 = (y-k)/2 = z/1 = mu`

`=> 2 lambda +1 = mu +3`;

`3 lambda -1 = 2 mu +k ; 4 lambda +1= mu`

`=>k = 9/2`
Correct Answer is `=>` (B) `9/2`
Q 2612245130

Find the value of `λ` so that the lines ,
` (1-x)/3 = (y-2)/(2 lamda) = (z-3)/2` and ` (x-1)/(3 lamda) = (y - 1)/1 = (6 - z)/7` are perpendicular to each other.
CBSE-12th 2009
Solution:

Given lines are

` (1-x)/3 = (y-2)/(2 lamda) = (z-3)/2`

and

` (x-1)/(3 lamda) = (y - 1)/1 = (6 - z)/7`

Let us rewrite the equations of the given lines as follows:

` (-(x-1))/3 = (y - 2)/(2lamda) = (z-3)/2`

and

` (x-1)/(3 lamda) = (y - 1)/1 = (-(z-6))/7`

That is we have,

` (x-1)/(-3) = (y - 2)/(2 lamda) = (z - 3)/2`

and

` (x-1)/(3 lamda) = (y -1)/1 = (z- 6)/(-7)`

The lines are perpendicular so angle between them is  `90^o`

So, `cos theta = 0`

Here `(a1,b1,c1)=(-3,2 lamda ,2)` and `(a2,b2,c2) = (3 lamda ,1,-7)`

For perpendicular lines

`a_1a_2 + b_1b_2 + c_1c_2 = 0`

` => - 9 lamda + 2 lamda - 14 = 0`

` => - 7 lamda - 14 =0`

` => lamda = (14)/(-7)`

` => lamda = -2`
Q 2559378214

The foot of the perpendicular from `P(1, 0, 2)` to the line `(x + 1)/3 = (y - 2)/(-2) = (z + 1)/(-1)` is the point

BCECE Mains 2015
(A)

`(1, 2, - 3) `

(B)

` ( 1/2 , 1 , -3/2)`

(C)

`(2, 4, -6)`

(D)

`(2, 3, 6)`

Solution:

`(x + 1)/3 = (y - 2)/(-2) = (z + 1)/(-1) = r`

`:. M = ( 3r - 1 , -2r + 2 , -r - 1)`

DR's of PM are `3r - 2,- 2r + 2,- r - 3`

DR's of line QM are `3, - 2, -1`.

`:. 9r - 6 + 4r - 4 + r + 3 = 0`

`=> 14r - 7 = 0`

`=> r = 1//2`

`:. M = ( 1/2 , 1 , -3/2)`
Correct Answer is `=>` (B) ` ( 1/2 , 1 , -3/2)`
Q 2305580468

Foot of perpendicular of point `(2, 2, 2)`

in the plane `x + y + z = 9` is
BITSAT Mock
(A)

`(1, 1, 1)`

(B)

`(3, 3, 3)`

(C)

`(9, 0, 0)`

(D)

`(2, 6, 1)`

Solution:

Equation of line perpendicular to
the plane through `(2, 2, 2)` is

`(x-2)/1 = (y-2)/1 = (z-2)/1 = 2`

`=> ( lambda + 2, lambda +2, lambda +2)` is a general point on it.

Now, `lambda +2 + lambda +2 + lambda + 2 =9`

`=> lambda = 1`

`:.` foot of the perpendicular is `(3, 3, 3)`.
Correct Answer is `=>` (B) `(3, 3, 3)`
Q 2459180914

The angle between the line `(3x -1) /3 = (y +3 )/(-1) = (5 -2z) /4` and the plane 3x- 3y - 6z =10` is equal to
BCECE Stage 1 2014
(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

None of these

Solution:

Given line can be rewritten as

`(x -1/3)/1 = (y +3)/(-1) = (z-5/2)/(-2)`

and equation of plane is written as

`x-y - 2z = 10/3`

Here ,direction ratios of line and plane are

`a_1 = 1, b_1 = - 1` and `c_1 = - 2`

and `a_2 = 1, b_2 = - 1` and `c_2 = - 2`

Now , `sin theta = (a_1a_2 + b_1b_2 + c_1c_2)/(sqrt(a_1^2 + b_1^2 +c_1^2 ) sqrt (a_2^2 + b_2^2 +c_2^2 )`

`=(1 xx 1 + (-1) xx (-1) + (-2) xx (-2))/(sqrt(1 + 1+ 4) sqrt (1 +1 +4 ))`

`=(1 + 1 + 4)/(sqrt6 sqrt 6) = 6/6 =1`

`=> theta = pi/2`
Correct Answer is `=>` (D) None of these
Q 2349191013

If `theta` is the acute angle between the diagonals of a
cube, then which one of the following is correct?
NDA Paper 1 2008
(A)

`theta < 30^(circ)`

(B)

`theta = 60^(circ)`

(C)

` 30^(circ) < theta < 60^(circ)`

(D)

`theta > 60^(circ)`

Solution:

We know that, the angle between the diagonals of

cube is `theta = cos^(-1) (1/3)`

If `theta` is an acute angle, then `theta > 60^(circ)`.
Correct Answer is `=>` (D) `theta > 60^(circ)`
Q 2379278116

What is the angle between the lines `x +z =0`

`y =0` and `20x=15y =12z`?
NDA Paper 1 2009
(A)

`cos^(-1) (1/5)`

(B)

`cos^(-1) (1/7)`

(C)

`sin^(-1) (1/5)`

(D)

`sin^(-1) (1/7)`

Solution:

Given that,


`x+z =0 ` ` y =0` and `20x =15 y =12z`

`=> x/1 = y/0 = z/-1` and `x/3 = y/4 = z/5`

Let `theta` be the angle between both lines.

Then, `cos theta = | ((1) (3) + (0) (4) + (-1) (5))/ (sqrt (1+0 +1) sqrt (9+16 +25)) | `

`= | (3+0 -5)/(sqrt(2) sqrt (50)) | = 2/(sqrt(2) * 5 sqrt (2)) = 1/5`

`:. theta = cos^(-1) (1/5)`
Correct Answer is `=>` (A) `cos^(-1) (1/5)`
Q 2329167911

What is the value of `n`, so that the angle between
the lines having direction ratios `(1, 1, 1)` and
`( 1, -1, n)` is `60^(circ)`?
NDA Paper 1 2009
(A)

`sqrt(3)`

(B)

`sqrt(6)`

(C)

`3`

(D)

None of these

Solution:

`:. cos 60^(circ) = | (1 xx 1 + 1 xx (-1) +1 xx n)/(sqrt (1^2 + 1^2 + 1^2 xx sqrt (1^2 + 1^2 + n^2) ) ) | `

`( :. cos theta = | (a_1 a_2 + b_1b_2 + c_1 c_2)/(sqrt (a_1^2 + b_1^2 + c_1^2 ) sqrt (a_2^2 + b_2^2 + c_2^2)) | )`

`=> 1/2 = n/(sqrt(3) sqrt (2+n^2) ) => 3 (2+n^2) = 4n^2`

`=> n^2 = 6 => n = pm sqrt(6) => n = sqrt(6)`
Correct Answer is `=>` (B) `sqrt(6)`
Q 2449780613

The angle between the lines whose direction cosines are `(sqrt3/4 , 1/4 , sqrt3/2)`and `(sqrt3/4 , 1/4 , - sqrt3/2)` is
BCECE Stage 1 2012
(A)

`pi`

(B)

`pi/2`

(C)

`pi/3`

(D)

`pi/4`

Solution:

`costheta = |l_1l_2+m_1m_2+n_1n_2|`


` = |sqrt3/4xxsqrt3/4+1/4xx1/4+sqrt3/2xx(-sqrt3/2)|`


` = |3/16+1/16-3/4| = |-2/4| = 1/2`


`=> theta = pi/3`
Correct Answer is `=>` (C) `pi/3`

Equation of plane and angel between two plane

Q 1638380202

A triangular plane `ABC`
with centroid `(1, 2, 3)` cuts the coordinate axes at `A, B, C`,
respectively

What is the equation of the plane `ABC?`
NDA Paper 1 2015
(A)

`x+2y+3z=1`

(B)

`3x+2y+z=3`

(C)

`2x+3y+6z=18`

(D)

`6x+3y+2z=18`

Solution:

We have,`x/ A + y/B + z/C = 1`

Since, triangular plane `ABC` with centroid `(1, 2, 3)` cuts

the coordinate axes, :. Intercepts made by

`:. X`-axis, `Y`-axis and `Z`-axis are, `3, 6, 9`.

`:.` Equation of the plane `ABC` is

`x/3 + y/6 + z/9 = 1` or `6x + 3y + 2z = 18`
Correct Answer is `=>` (D) `6x+3y+2z=18`
Q 1618480300

A point `P(1, 2, 3)` is one
vertex of a cuboid formed by the coordinate planes and the
planes passing through `P` and parallel to the coordinate
planes.

What is the equation of the plane passing through
`P(1, 2, 3)` and parallel to `xy` -plane?
NDA Paper 1 2015
(A)

`x + y = 3`

(B)

`x - y = -1`

(C)

`z = 3`

(D)

`x+2y+3z=14`

Solution:

The equation of the plane passing through

`P (1, 2, 3)` and parallel to `xy` - plane is `z = 3`.
Correct Answer is `=>` (C) `z = 3`
Q 1629591411

Consider the plane passing through the points `A(2, 2, 1),
8(3, 4, 2)` and `C(7, 0, 6)`.

Which one of the following points lies on the plane?
NDA Paper 1 2014
(A)

`(1, 0, 0)`

(B)

`(1, 0, 1)`

(C)

`(0, 0, 1)`

(D)

None of these

Solution:

We know that, equation of plane passing

through three non-collinear points `(x_1, y_1, z_1), (x_2 , y_2, z_2)`

and `(x_3 , y_3 , z_3 )` is

` |(x - x_1, y - y_1, z - z_1),(x_2 - x_1 , y_ 2 - y_1, z_2 - z_1),( x_3 - x_1, y_3 - y_1, z_3 - z_1)| = 0`

Put the value of `(x_1, y_1, z_1), (x_2 , y_2, z_2 )` and `(x_3 , y_3, z_3 )`, we get

` |(x-2, y-2, z-1),(1, 2 ,1) ,(5, -2, 5)| = 0`

`=> (x -2)(10 + 2) - (y -2)(5- 5) + (z -1) (-2 -10) = 0`

`=> 12x - 12z =12 => x - z = 1`

Hence, the equation of plane passes through `(1, 0, 1 )`.
Correct Answer is `=>` (B) `(1, 0, 1)`
Q 2319123910

What is the angle between the planes

`2x - y - 2z + l = 0` and `3x - 4y + 5z - 3 = 0`?
NDA Paper 1 2013
(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

`pi/2`

Solution:

Given equation of planes


`2x-y-2z+1=0` ... (i)

and `3x-4y+ 5z-3 = 0 `... (ii)

Here, `a_1=2,b_1 =-1, c_1=-2` and `a_2 =3,b_2 =-4,c_2= 5` are

direction ratios of planes Eqs. (i) and (ii).

Let `theta` be the angle between two planes.

Then , `cos theta = ( |a a_2 +b_1 b_2 + c_1 c_2 | )/(sqrt ( a_1^2 +b_1^2 + c_1^2) sqrt (a_2^2 +b_2^2 +c_2^2) )`

`= ( | (2)(3) + (-1) (-4) + (-2) (5) | )/(sqrt (4+1+4) sqrt (4+16 +25) ) = (|6+4 -10 | )/(sqrt (9) sqrt (45)) =0 `

`= cos pi/2`

So, the required angle is `pi/2`
Correct Answer is `=>` (D) `pi/2`
Q 2369834715

What is the equation to the plane through
`(1, 2, 3)` parallel to `3x + 4y -5z =0`?
NDA Paper 1 2012
(A)

`3x + 4y + 5z + 4 = 0`

(B)

`3x + 4y- 5z + 14 = 0`

(C)

`3x + 4y -· 5z + 4 = 0`

(D)

`3x + 4y- 5z- 4 = 0`

Solution:

The equation of any plane parallel to the plane

`3x + 4y- 5z = 0` may be taken as

`3x + 4y- 5z + k = 0` ... (i)

If plane (i) passes through the point `(1, 2, 3)`, then

`3(1) + 4(2)- 5(3) + k = 0`

`=> 3 + 8-15 + k = 0`

`=> 4+k= 0 => k=4` ... (ii)

On putting `k = 4` in Eq. (i), we get required equation

i.e., `3x + 4y- 5z + 4 = 0`
Correct Answer is `=>` (C) `3x + 4y -· 5z + 4 = 0`
Q 2319045810

What is the equation of the plane passing through
the point `(1, -1, -1)` and perpendicular to each of
the planes `x- 2y - 8z = 0` and `2x + 5y- z = 0?`
NDA Paper 1 2011
(A)

`7x- 3y+ 2z= 14`

(B)

`2x + 5y- 3z = 12`

(C)

`x - 7 y + 3z = 4`

(D)

`14x- 5y+ 3z= 16`

Solution:

Let the equation of the plane which passes through the
point `(1, -1, - 1)` be,

`a(x-1)+b(y+1)+c(z+1)=0` ... (i)

where, `a, b` and `c` are the direction ratios of the normal of the
plane (i).

Now, plane (i) is perpendicular to the planes `x - 2 y - 8z = 0` and

`2x + 5y -z =0`,then

`a-2b -8c =0`..........(ii)

and `2a + 5b -c =0` ..............(iii)

Now, solving Eqs. (ii) and (iii) by cross-multiplication method,

`a/(2+40) = b/(-16 +1) = c/(5+4)`

`=> a/42 = b/-15 = c/9`

From Eq. (i),

`42 (x - 1)- 15(y + 1) + 9(z + 1) = 0`

`=> 42x-15y+9z=42+15-9`

`=> 42x-15y+9z=48`

`=> 14x- 5y + 3z = 16`
Correct Answer is `=>` (D) `14x- 5y+ 3z= 16`
Q 2319145910

What is the angle between the planes

`2x - y + z = 4` and `x + y + 2z = 6` ?
NDA Paper 1 2011
(A)

`pi/2`

(B)

`pi/3`

(C)

`pi/4`

(D)

`pi/6`

Solution:

Let the angle between two planes `2x- y + z = 4` and

`x + y + 2 z = 6` be `theta`, then

`cos theta = | (a_1 a_2 + b_1 b_2 +c_1 c_2)/(sqrt (a_1^2 + b_1^2 + c_1^2) sqrt (a_2^2 +b_2^2 +c_2^2) ) |`

where, `a_1 = 2, b_1 = -1,c_1 = 1` and `a_2 = 1,b_2 = 1` and `c_2 = 2`

Then, `cos theta = | ( (2)(1) + (-1) (1) + (1) (2))/(sqrt (4+1+1) sqrt (1+1+4) ) |`

`=> cos theta = |(2-1 +2)/(sqrt(6) sqrt (6)) | = |3/6 | = 1/2`

`=> cos theta = cos pi/3`

`:. theta = pi/3`
Correct Answer is `=>` (B) `pi/3`
Q 2239191012

Equation of the plane containing the

line `(x-3)/2 = (y-4)/3 = (z-5)/4` is
BITSAT Mock
(A)

`4x + 4y - 5z = 3`

(B)

`2x + 3y - 4z + 5 = 0`

(C)

`4x + 6y + 8z + 13 = 0`

(D)

`x + 2y + z = 0`

Solution:

Correct Answer is `=>` (A) `4x + 4y - 5z = 3`
Q 1655823764

The equation of the plane through the intersection of the
planes `x + y + z = 1` and `2x + 3 y - z + 4 = 0` parallel to
`x-` axis, is
BITSAT 2016
(A)

`y - 3z + 6 = 0`

(B)

`3y - z + 6 = 0`

(C)

`y + 3z + 6 = 0`

(D)

`3y- 2z + 6 = 0`

Solution:

The equation of the plane through the intersection of the]

planes `x + y + z = 1 ` and `2x + 3y - z + 4 = 0`

`(x + y + z -1) +A (2x + 3y- z + 4) = 0` is

as, `(2lamda + 1) x + (3lamda + 1)y + (1 - lamda) z + 4lamda -1 = 0` ........(i)

It is parallel to x-axis i.e. ` x/1 = y/0 = z/0`

`: . 1 (2lamda + 1) + 0 X (3lamda + 1) + 0 (1-lamda) = 0`

` => lamda = -1/2`

Substituting `lamda = -1/2` in Eq. (i), we get

`y- 3z + 6 = 0` as the equation of the required plane.
Correct Answer is `=>` (A) `y - 3z + 6 = 0`
Q 1580878717

The image of the point `(3, 2, 1)` in the plane `2x-y+3z = 7` is
BITSAT 2009
(A)

`(1, 2, 3)`

(B)

`(2, 3, 1)`

(C)

`(3, 2, 1)`

(D)

`(2, 1, 3)`

Solution:

We know that image `(x, y, z)` of a point `(x_1, y_1, z_1)` in a plane `ax + by + cz + d = 0` is

` (x-x_1)/a = (y -y_1) /b = (z -z_1) /c`

` = (-2 ax_1 + by_1 + cz_1 + d)/(a^2 +b^2 + c^2 )`

Here, point is `(3, 2, 1)` and plane is `2x - y + 3z = 7`.

` :. ( x-3)/2 = ( y-2)/-1 = ( z-1)/3`

` ( -2quad 2quad 3 -2 + 3quad 1 -7)/ (2^2 + 1^2 + 3^2)`

`=> (x-3)/2 = (y-2)/(-1) = ( z-1)/3= -2 quad 0`

`=> x = 3 , y =2 , z =1 `
Correct Answer is `=>` (C) `(3, 2, 1)`
Q 1521134921

The equation of plane passing through a point `A (2, -1, 3)` and parallel to the vectors

`vec a = (3, 0,- 1)` and `vecb = (-3, 2, 2)` is :
BITSAT 2005
(A)

`2x - 3y + 6z - 25 = 0`

(B)

`2x-3y + 6z + 25 = 0`

(C)

`3x - 2y + 6z- 25= 0`

(D)

`3x - 2y + 6z + 25 = 0`

Solution:

The equation of any plane through `(2, -1, 3)` is

`a (x - 2) + b (y + 1) + c (z - 3) = 0` ... (i)

where `a, b` and `c` are direction ratios, Since

Eq. (i) Is parallel to `vec a` and `vec b`

`:. 3a+ 0b- c= 0` ... (ii)

and `- 3a + 2b - 2c = 0` ... (iii)

Solving Eqs. (ii) and (iii), we get

`a/2 =-b/(6-3) =c/6 =k` say

`=> a= 2k, b =-3k, c =6k`

Putting the values of `a, b` and `c` in Eq. (i), we get
.
`2k (x - 2)- 3k (y + 1) + 6k (z- 3) = 0`.

`=> 2x - 3y + 6z- 25 = 0`

which is a required equation of a plane.
Correct Answer is `=>` (A) `2x - 3y + 6z - 25 = 0`
Q 2816412379

The equation of the plane passing through the
point (-2,- 2, 2) and containing the line joining
the points (1, 1, 1) and (1,- 1, 2) is

(A)

x+2y- 3z+4=0

(B)

3x-4y+1= 0

(C)

5x+2y-3z - 17= 0

(D)

x-3y -6z + 8 = 0

Solution:

Equation of a plane through
`( -2,- 2, 2)` is given by

`a(x + 2) + b (y + 2) + c (z- 2) = 0`
It contains the line joining the points
`(1, 1, 1)` and `B (1, 1, 2)`, so these
points also lie in the plane.

`:.` At `(1, 1, 1) 3a + 3b- c = 0`

and at `(1, -1 ,2) 3a + b + 0 * c = 0`

`=> a/1 = b/(-3) = c/(-6) = r` (say)

`=> a = r , b = -3 r, c= - 6 r`

So, equation of the plane is

`x - 3 y - 6z + 8= 0`.
Correct Answer is `=>` (D) x-3y -6z + 8 = 0
Q 2480501417

Which one of the following is correct'?
The three planes `2x + 3y -z -2 =0`,
`3x + 3 y + z - 4 = 0` and `x - y + 2z - 5 = 0`
intersect
NDA Paper 1 2007
(A)

at a point

(B)

at two points

(C)

at three points

(D)

in a line

Solution:

We know that, planes never intersect at points, they
intersect in a line.
Correct Answer is `=>` (D) in a line
Q 2400701618

Which one of the following is the plane

containing the line `(x-2)/2 = (y-3)/3 = (z-4)/5` are parallel to `Z`-axis?
NDA Paper 1 2007
(A)

`2x- 3y = 0`

(B)

`5x- 2z = 0`

(C)

`5y- 3z = 0`

(D)

`3x -2y= 0`

Solution:

The equation of the plane which is parallel to `Z`-axis is

`ax+by+d=0`.

Since, the line is passing through the point `(2, 3, 4)`, then it will
satisfy the equation of required plane.

From option (d) `3x- 2y = 3(2)- 2(3) = 6- 6 = 0`
Correct Answer is `=>` (D) `3x -2y= 0`
Q 2379691516

What is the equation of the plane pa&sing through
`(x_1, y_1, z_1)` and normal to the line with `< a, b, c >` as
direction ratios?
NDA Paper 1 2007
(A)

`ax + by+ cz = ax_1 + by_1 + cz_1`

(B)

`a(x + x_1)+ b(y+ y_1)+ c(z+ z_1)= 0`

(C)

`ax + by+ cz = 0`

(D)

`ax+ by+ cz = x_1 + y_1 + z_1 = 0`

Solution:

The equation of the plane passing through `(x_1, y_1, z_1)`
and normal to the line with `< a, b, c >` as direction ratios, is

`a(x-x_1)+b(y-y_1)+c(z-z_1)=0`

`=> ax - ax_1 + by- by_1 + cz - cz_1 = 0`

`=> ax + by + cz = ax_1 + by_1 + cz_1`

which is required equation of plane.
Correct Answer is `=>` (A) `ax + by+ cz = ax_1 + by_1 + cz_1`
Q 2379191016

Which one of the following planes contains the
`Z`-axis?
NDA Paper 1 2008
(A)

`x- z = 0`

(B)

`z + y = ()`

(C)

`3x + 2y = 0`

(D)

`3x + 2z = 0`

Solution:

The equation of plane which contains `Z`-axis is

`3x + 2 y = 0`. (by property)
Correct Answer is `=>` (C) `3x + 2y = 0`
Q 2329067811

What is the equation of the plane through `Z`-axis

and parallel to the line `(x-1)/(cos theta) = (y+2)/(sin theta) = (z-3)/0 `
NDA Paper 1 2010
(A)

`x cot theta + y = 0`

(B)

`x tan theta - y = 0`

(C)

`x + y cot theta = 0`

(D)

`x-y tan theta = 0`

Solution:

Equation of plane through `Z`-axis is

`ax + by = 0` ..............(i)

This plane is parallel to the line

`(x-1)/(cos theta) = (y+2)/(sin theta) = (z-3)/0`

`=> a cos theta + b sin theta = 0 => a= -b tan theta`

`:. -b tan theta x + by = 0 => x tan theta -y =0`

Which is the required equation of plane.
Correct Answer is `=>` (B) `x tan theta - y = 0`
Q 2319780610

Under which one of the following conditions will
the two planes `x + y + z = 7` and `alpha x + beta y + gamma z = 3`,
be parallel (but not coincident)?
NDA Paper 1 2008
(A)

`alpha = beta = gamma =1`

(B)

`alpha = beta = gamma = 3/7`

(C)

`alpha = beta = gamma`

(D)

None of these

Solution:

The given equation of planes are

`x + y + z = 7` .................(i)

and `alpha x + beta y + gamma z = 3`..................(ii)

These planes will be parallel, if the coefficients of `x, y` and `z` are
the same, i.e ..

`alpha = beta =gamma`
Correct Answer is `=>` (C) `alpha = beta = gamma`
Q 2379378216

The equation `by + cz + d = 0` represents a plane
parallel to which one of the following?
NDA Paper 1 2008
(A)

`X`-axis

(B)

`Y`-axis

(C)

`Z`-axis

(D)

None of these

Solution:

The equation of plane which is parallel to `X`-axis is
`by + cz = d`, where `x = 0`.
Correct Answer is `=>` (A) `X`-axis
Q 2319480319

Which one of the following planes is normal to
the plane `3x + y + z = 5`?
NDA Paper 1 2008
(A)

`x+ 2y + z = 6`

(B)

`x- 2y + z = 6`

(C)

`x + 2y- z = 6`

(D)

`x- 2y- z = 6`

Solution:

We know that, the condition of perpendicularity of two
planes is

`a_1a_2 + b_1b_2 + c_1 c_2 = 0`

where, `a_1, b_1, c_1` and `a_2, b_2, c_2` are the direction ratios of the
normal to the planes, respectively.

Given plane is `3x + y + z = 5`.

From option `(d), x- 2y- z = 6`

So, `(3)(1) + (1)(-2) + (1)(-1) = 3- 2 - 1 = 0`
Correct Answer is `=>` (D) `x- 2y- z = 6`
Q 2319278119

What is the equation of a plane through the `X`-axis
and passing through the point `(1, 2, 3)`?
NDA Paper 1 2009
(A)

`x + y + z = 6`

(B)

`x=1`

(C)

` y+z =5`

(D)

`z+y =1`

Solution:

The equation of the plane passing through `X`axis is

`x = a`.

This also passes through `(1, 2, 3)`

`:. x =1` `( :. a =1 )`

which is the required equation of plane
Correct Answer is `=>` (B) `x=1`
Q 2308312208

What is the number of planes passing through
three non-collinear points?
NDA Paper 1 2009
(A)

`3`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

We know that, the number of planes passing through
three non-collinear points is `1` .
Correct Answer is `=>` (C) `1`
Q 2309278118

What is the angle between the planes

`2x - y +z =6` and `x+y +2z =3`?
NDA Paper 1 2009
(A)

`pi/2`

(B)

`pi/3`

(C)

`pi/4`

(D)

`pi/6`

Solution:

Let `theta` be the angle between given planes, then

`cos theta = | (2 xx 1 + 1 xx (-1) + 1 xx 2)/(sqrt (4+1+1) sqrt (1+1+4)) |`

`= 3/6 =1/2 = cos pi/3`

`=> theta = pi/3`
Correct Answer is `=>` (B) `pi/3`
Q 2359167914

The direction cosines of a line are proportional to
`(2, 1, 2)` and the line intersects a plane
perpendicularly at the point `( 1,- 2, 4)`. What i~ the
distance of the plane from the point `(3, 2, 3)`?
NDA Paper 1 2009
(A)

`sqrt(3)`

(B)

`2`

(C)

`2 sqrt(2)`

(D)

`4`

Solution:

Equation of plane passing through `(1, -2, 4)` and the
direction cosines of whose normal `(2, 1, 2 )`, is

`2(x-1)+ 1(y+2)+2(z-4)=0`

`=> 2x+y+2z-8=0`

`:.` Required distance

`= | (2(3) + 1(2) +2(3) - 8)/(sqrt (4+1+4) ) | ` `( :. ` distance `= | (ax+by +c)/(sqrt (a^2 + b^2) )| `)

`= 6/3 = 2`
Correct Answer is `=>` (B) `2`
Q 2389156917

Which one of the following points lies on the
plane `2x + 3y- 6z =21`?
NDA Paper 1 2011
(A)

`(3, 2, 2)`

(B)

`(3, 7, 1)`

(C)

`(1, 2. 3)`

(D)

`(2, 1, -1)`

Solution:

Since, the point `(3, 7, 1)` satisfies the equation of plane

`2x + 3y- 6z = 21`

Hence, `(3, 7, 1)` lies on the plane.
Correct Answer is `=>` (B) `(3, 7, 1)`
Q 2389467317

If the planes `px + 2 y + 2z - 3 = 0` and

`2x - y + z + 2 = 0` intersect at an angle `pi/4` , then

what is the value of `p^2` ?
NDA Paper 1 2010
(A)

`24`

(B)

`12`

(C)

`6`

(D)

`3`

Solution:

We know that, the angle between the planes
`a_1x + b_1y + c_1z + d_1 = 0` and `a_2x + b_2y + c_2z + d_2 = 0` is given by


`cos theta = | (a_1a_2 + b_1b_2 +c_1 c_2)/(sqrt (a_1^2 + b_1^2 +c_1^2) sqrt (a_2^2 +b_2^2+c_2^2)) |`

` :. cos pi/4 = | (p xx 2 +2 xx (-1) +2 xx 1)/(sqrt (p^2 +4+4) sqrt (4+1+1) ) |`

`=> 1/(sqrt(2)) = (2p)/(sqrt (p^2 + 8 sqrt (6)) )`

`=> 1/2 = (4p^2)/(( p^2 + 8) sqrt(6))`

`=> 1/2 = (4p^2)/( (p^2 +8) 6 ) => p^2 = 24`

The vertices of a cube are `(0, 0, 0), (2, 0, 0), (0, 2, 0)`,
`(0, 0, 2), (2, 2, 0)`, and `(2, 0, 2), (0, 2, 2), I (2, 2, 2)`,
respectively.
Correct Answer is `=>` (A) `24`
Q 2319167010

The two planes `ax + by + cz + d = 0` and
`ax+ by + cz + d_1 = 0`, (where, `d ne d_1)` have
NDA Paper 1 2010
(A)

one point only in common

(B)

three points in common

(C)

infinite points in common

(D)

no points in common

Solution:

Since, two planes `ax + by+ cz + d = 0` and
`ax+ by+ cz + d_1 = 0` are parallel to each other. `( :. d ne d_1 )`


So, there have no common point.
Correct Answer is `=>` (D) no points in common
Q 2826001871

The line `(x-2)/3 = (y-3)/4 = (z-4)/5` is parallel to the plane

(A)

`2x + y - 2z = 0`

(B)

`3x + 4y + 5z = 7`

(C)

`x+ y+ z =2`

(D)

`2x + 3y + 4z = 0`

Solution:

Let the line `(x- alpha )/l = (y- beta)/m = (z- gamma)/n` is

parallel to the plane

`ax + by + cz + d = 0`

Then, normal to the plane is
perpendicular to the line.
i.e. `al + bm + cn= 0`

In this question, this condition is
satisfied by the plane `2x + y - 2z = 0`.
Correct Answer is `=>` (A) `2x + y - 2z = 0`
Q 2369856715

What is the acute angle between the planes

`x + y + 2z = 3` and `-2x + y - z = 11`?
NDA Paper 1 2011
(A)

`pi/5`

(B)

`pi/4`

(C)

`pi/6`

(D)

`pi/3`

Solution:

Since, the equation of planes are `x + y + 2 z = 0` and

`-2x + y- z = 11`.

We know tllat, the angle betweep the planes

`a_1 x + b_1 y +c_1 z + d_1 =0` and `a_2 x + b_2 y + c_2 z + d_2 =0` is

given by

`cos theta = | (a_1 a_2 + b_1 b_2 + c_1 c_2)/ (sqrt (a_1^2 + b_1^2 + c_1^2) sqrt (a_2^2 +b_2^2 + c_2^2) ) |`

Here, `a_1 = 1 , b_1 = 1,c_1 =2` and `a_2 = -2, b_2 =1, c_2 = -1`

`:. cos theta = | (1 xx (-2) +1 xx 1 +2 xx (-1) )/ (sqrt (1+1+4) sqrt (4+1+1)) |`

`= |(-2+1-2)/(sqrt(6)sqrt (6)) | = 3/6 =1/2 = cos\ \ pi/3`

`=> theta = pi/3`
Correct Answer is `=>` (D) `pi/3`

Intersection of line and plane : Angel between a line & plane, coplanarity of two line

Q 2763291145

The line passing through the points `(1, 2, - 1)` and `(3, - 1, 2)` meets the yz-plane at which one of the following points?
NDA Paper 1 2017
(A)

`( 0 , -7/2 , 5/2)`

(B)

`( 0 , 7/2 , 1/2 )`

(C)

`( 0 , -7/2 , -5/2)`

(D)

`( 0 , 7/2 , -5/2)`

Solution:

Line passing through the given point

`(x-1)/2 = (y-2)/(-3) =(2+1)/3`

For `yz` plane `x=0`

`(y-2)/(-3) =(-1)/2`

`y= 7/2`

`z= (-5)/2`
Correct Answer is `=>` (D) `( 0 , 7/2 , -5/2)`
Q 2167123985

A plane P passes through the line of intersection of the planes
`2x- y + 3z = 2 , x + y - z = 1` and the point `(1, 0, 1).`

What are the direction ratios of the line of intersection of
the given planes?
NDA Paper 1 2016
(A)

`< 2, - 5, 3 >`

(B)

`< 1, - 5, - 3 >`

(C)

`< 2,5, 3>`

(D)

`< 1, 3,5 >`

Solution:

Given equations of plane are

`2x- y + 3z ,= 2` ............(i)

and `x + y - z = 1` ............(ii)

Let the direction ratios of line of intersection be

`< a, b, c >.`

`:. 2a - b + 3c = 0`...... ... (iii)

and `a + b - c = 0 [∵ a_(1) a_(2) + b_(1) b_(2) + c_(1) c_(2) = 0]` ........ (iv)

On solving Eqs. (iii) and (iv), we get

` a/-2 = (-b) /-5 = c/3 `

` => a/-2 = b/5 = c/5 ` or ` a/2 = b/-5 = c/ -3 `

` => < a, b, c > = < 2, -5, -3 >`
Correct Answer is `=>` (A) `< 2, - 5, 3 >`
Q 2177134086

A plane P passes through the line of intersection of the planes
`2x- y + 3z = 2 , x + y - z = 1` and the point `(1, 0, 1).`

What is the equation of the plane P?
NDA Paper 1 2016
(A)

`2x+5y-2=0`

(B)

`5x+ 2y -5=0`

(C)

`x + z- 2 = 0`

(D)

`2x- y - 2z = 0`

Solution:

The equation of plane `P` is

`(2x - y + 3z - 2) + lamda (x + y - z - 1) = 0`

It passes through `(1, 0, 1)`.

We get, `(2- 0 + 3 -2 ) lamda (1 + 0 -1 - 1) = 0`

`=> 3 - lamda =0 => lamda = 3`

Hence, equation of plane `P` is

`(2x- y + 3z- 2) + 3(x + y- z -1) = 0`

`=> 5x + 2y - 5 = 0`
Correct Answer is `=>` (B) `5x+ 2y -5=0`
Q 2251834724

The lengths of the intercepts on the coordinate axes
made by the plane `5x + 2y + z -13 = 0` are
NDA Paper 1 2015
(A)

`(5, 2, 1)` units

(B)

`( (13)/5, ( 13)/2 ,13)` units

(C)

`( 5/(13) , 2 /(13) , 1/(13))` units

(D)

`(1, 2, 5)` units

Solution:

Given `5x + 2y + z -13 = 0` or ` x/(13//5) + y/ (13//2) =z/(13)= 1`

`:.` Lengths of intercepts are `(13)/5 , (13)/2 ` and `13`.
Correct Answer is `=>` (B) `( (13)/5, ( 13)/2 ,13)` units
Q 1619491319

The line joining the points `(2, 1, 3)` and `(4, -2, 5)` cuts the
plane `2x + y - z = 3.`

Where does the line cut the plane?
NDA Paper 1 2014
(A)

`(0, - 4, -1)`

(B)

`(0, - 4, 1)`

(C)

`(1, 4, 0)`

(D)

`(0, 4, 1)`

Solution:

Equation of line passing through the points

`(2, 1, 3)` and `(4, -2, 5)` is

`(x-2)/(4-2) = (y-1)/(-2-1) = (z-3)/(5-3) = lamda`

` => (x-2)/2 = (y-1)/(-3) = (z-3)/2 = lamda`

`=> x = 2lamda + 2, y = - 3lamda + 1` and `z = 2lamda + 3`

Since, this line cuts the plane `2x + y - z = 3`.

So, `(2lamda + 2, - 3lamda + 1, 2lamda + 3)` satisfies the equation of plane.

`:. 2lamda + 2, - 3lamda + 1 - 2lamda - 3 = 3`

`=> -3lamda = 3`

` => lamda = -1`

Hence, points are `[2(-1) + 2,- 3(-1) + 1, 2 (-1) + 3]`, i.e .

`(0, 4, 1)`.
Correct Answer is `=>` (D) `(0, 4, 1)`
Q 2475434366

The lines `(x-a +d )/(alpha - delta) = (y-a)/alpha = (z-a-d)/(alpha+ delta )` and

`(x-b +c)/(beta - r) = (y-b)/beta =(z-b -c)/(beta+ r)` are coplanar and

then equation to the plane in which they lie, is
UPSEE 2012
(A)

`x + y + z = 0`

(B)

`x- y + z = 0`

(C)

`x- 2y + z = 0`

(D)

`x + y - 2z = 0`

Solution:

The lines will be coplanar, if

`| (a-d-b+c, a-b, a+d-b -c), (alpha -delta , alpha , alpha +delta), (beta-r, beta , beta +r) | =0`


Add 3rd column to first and it becomes twice the
second and hence the determinant is zero, as
the two columns are identical. Again, the
equation of the plane in which they lie is

`| (x-a+d , y -a , z-a-d), (alpha -delta , alpha , alpha +delta), (beta -r, beta , beta +r) | = 0`

On adding 1st and 3rd columns and subtracting
twice the 2nd, we get

`| (x+z-2y, y-a ,z-a-d ), (0,alpha , alpha +delta) ,(0,beta, beta+r) | =0`

`=> [alpha (beta +r) - beta (alpha +delta) ] (x+z -2y) =0`

`=> (x+z-2y ) =0`
Correct Answer is `=>` (C) `x- 2y + z = 0`
Q 2570456316

The angle between the line `(x+1)/(2) = y/3 = (z -3)/6` and the plane `10 x + 2y - 11 z = 3` is
BCECE Stage 1 2013
(A)

`pi/2 `

(B)

`pi/4`

(C)

`pi/6`

(D)

`sin^(-1) (8/(21))`

Solution:

Let `theta` be the angle between the line and the

plane, we have

`a = 2, b = 3, c = 6`

and `a_1 = 10, b_2 =2, c_1 = -11`

`:. cos(90 - theta ) = sin theta`

` = | ( a a_1 + b b_1 + c c_1)/(sqrt(a^2 + b^2 + c^2) sqrt(a_1^2 + b_1^2 + c_1^2))|`

` = | ( (2)(10) + (3)(2)+ (6)(-11) )/( sqrt(2^2 + 3^2 + 6^2 )sqrt(10^2 + 2^2 + (-11)^2) )|`

` = | ( 20 + 6 - 66)/(sqrt (49) . sqrt (225))| = | (-40)/(7.15)|`

` theta = sin^(-1) ( 8/(21))`
Correct Answer is `=>` (D) `sin^(-1) (8/(21))`
Q 2369134015

If the straight line `(x-x_0)/l = (y-y_0)/m = (z-z_0)/n` is

parallel to the plane `ax + by + cz + d = 0`, then
which one of the following is correct?
NDA Paper 1 2013
(A)

`I + m + n = 0`

(B)

`a + b + c = 0`

(C)

`a/l + b/m +c/n =0`

(D)

`al + bm + cn = 0`

Solution:

Given that, equation of straight line is

`(x-x_0)/l = (y-y_0)/m = (z-z_0)/n` ................(i)

and equation of plane is

`ax + by +cz +d =0` .............(ii)

Since, the straight line is parallel to the plane i.e., normal to the
plane is perpendicular to the straight line.

By perpendicularity condition,

`l_1 l_2+m_1 m_2 + n_1 n_2 = 0`

`=> al + bm + cn = 0`
Correct Answer is `=>` (D) `al + bm + cn = 0`
Q 2439645512

A straight line joining the points `(1, 1, 1)` and
`(0, 0, 0)` intersects the plane `2x + 2y + z = 10`
at
WBJEE 2016
(A)

`(1, 2, 5)`

(B)

`(2, 2, 2)`

(C)

`(2, 1, 5)`

(D)

`(1, 1, 6)`

Solution:

Fquation of line joining the points `(1, 1, 1)` and
`(0 ,0, 0)` is

`(x-0)/(1-0) = (y-0)/(1-0) = (z-0)/(1-0) = lambda` (say)

`=> x=y =z= lambda`

So, the point is `(lambda , lambda, lambda)`

The point intersects the plane `2x + 2y + z = 10`.

`:. 2(lambda) + 2(lambda) +lambda= 10`

` => 5 lambda = 10`

`=> lambda =2`

Hence, the point is `(2, 2, 2)`.
Correct Answer is `=>` (B) `(2, 2, 2)`
Q 2259656514

The position vectors of point `A` and `B` are
` hat i − hat j + 3 hat k` and `3 hat i + 3 hat j + 3 hat k`
respectively. The equation of a plane is
` vec r . (5 hat i + 2 hat j − 7 hat k ) + 9 = 0`. The points
`A` and `B` :
BITSAT Mock
(A)

lie on the plane

(B)

lie on the same side of the plane

(C)

lie on the opposite side of the plane

(D)

None of these

Solution:

The position vectors of two given points

are

`vec a = hat i − hat j + 3 hat k` and `vec b = 3 hat i + 3 hat j + 3 hat k`

and the equation of the given plane is

` vec r . (5 hat i + 2 hat j − 7 hat k ) + 9 = 0`

or `vec r . vec n + d = 0`

We have

` vec a . n + d = ( hat i − hat j + 3 hat k ) . (5 hat i + 2 hat j − 7 hat k ) + 9`

`= 5 − 2 − 2 1 + 9 < 0`

and `vec b . vec n + d = (3 hat i + 3 hat j + 3 hat k )`

`. (5 hat i + 2 hat j − 7 hat k ) + 9`

`= 15 + 6 − 21 + 9`

` > 0`

So, the points `vec a ` and `vec b` are on the opposite

sides of the plane.
Correct Answer is `=>` (C) lie on the opposite side of the plane
Q 2418780609

The direction ratios of the normal to the
plane passing through the points `(1,- 2, 3),`
`( -1, 2, - 1)` and parallel to the line

`(x-2)/2 = (y+1)/3 = z/4` are proportional to
BCECE Stage 1 2015
(A)

`2, 3, 4`

(B)

`4, 0, 7`

(C)

`-2, 0,-1`

(D)

`2,0, -1`

Solution:

The equation of a plane passing through `(1, -2, 3)` is
`a(x-1)+b(y+2)+c(z-3)=0` ... (i)
It passes through `( -1, 2, -1)` and is parallel to the
given line.
`therefore a(-2) + b(4) + c(-4) = 0`
and `2a + 3b + 4c = 0`


`=> a/(28) = b/0 = c/(-14)`


`=> a/2 = b/0 = c/(-1)`

Hence, `a : b : c = 2 : 0 : - 1`
Q 1685780667

The lines `(x -1)/1 = (y-3)/1 = (z-4)/(-k)` and `(x-1)/k = (y-4)/2 = (z-5)/1` are coplaner, if
BITSAT 2016
(A)

`k = 3` or `- 3`

(B)

`k = 0` or `- 1`

(C)

`k = 1` or `-1`

(D)

`k = 0` or `-3`

Solution:

We know that the lines,

`(x-x_1)/l_1 = (y-y_1)/m_1 = (z-z_1)/n_1`

and `(x-x_2)/l_2 = (y-y_2)/m_2 = (z-z_2)/n_2`

are coplanar, iff

`|(x_2 -x_1 ,y_2 -y_1 ,z_2- z_1),( l_1, m_1, n_1),( l_2 ,m_2, n_2)| =0`

So, the given lines will be coplanar, if

`|(1-2, 4-3, 5-4),( 1, 1, -k), (k,2,1)| =0`

`=> k ^2 + 3k = 0 => k = 0, - 3`
Correct Answer is `=>` (D) `k = 0` or `-3`
Q 2309723618

What is the distance between the planes

`x - 2 y + z - 1 = 0` and `-3x + 6 y - 3z + 2 =0 ?`
NDA Paper 1 2013
(A)

3 units

(B)

1 unit

(C)

0

(D)

None of these

Solution:

The given equation of planes

`x-2y+ z-1 = 0` ................(i)

and `-3x+ 6y-3z + 2 = 0`

`=> x-2y+ z-2/3 = 0` ..............(ii)

Since, both planes are parallel to each other, then distance
between them

`= |-2/3 - (-1) |/(sqrt ((1)^2 + (-2)^2 + (1)^2 ) ) = | -2/3+1|/(sqrt (1+4+1) )= (1/3)/sqrt(6) =1/(3 sqrt (6))`
Correct Answer is `=>` (D) None of these

Shortest distance between two skew line

Q 1668380205

A point `P(1, 2, 3)` is one
vertex of a cuboid formed by the coordinate planes and the
planes passing through `P` and parallel to the coordinate
planes.

What is the length of one of the diagonals of the
cuboid?
NDA Paper 1 2015
(A)

` sqrt(10) units`

(B)

` sqrt(14) units`

(C)

`4 units`

(D)

`5 units`

Solution:

Length of diagonals of the cuboid

`= sqrt( 1^2 + 2^2 + 3^2) = sqrt(14) units`
Correct Answer is `=>` (B) ` sqrt(14) units`
Q 1865412365

Find the shortest distance between the lines
whose vector equations are
`vec r = (1 - t) hat i + (t - 2 ) hat j + (3 - 2t ) hat k` and
` vec r = (s + 1) hat i +(2s - 1) hat j - (2s + 1) hat k`
Class 12 Exercise 11.2 Q.No. 17
Solution:

The given equation

`vec r = hat i -2 hat j + 3 hat k + lamda (-hat i + hat j - 2 hat k)` .........(i)

`vec r = hat i - hat j - hat k + mu (-hat i + hat j - 2 hat k)` .........(ii)

Comparing these equation with ` vec r = vec a_1 + lamda vec b_1` and

` vec r = vec a_2 + mu vec b_2`

`:. vec a_1 = hat i -2 hat j + 3 hat k , vec b_1 = -hat i + hat j - 2 hat k` and

` vec a_1 = hat i - hat j - hat k , vec b_2 = hat i + 2 hat j - 2 hat k `

Now, ` vec a_2 - vec a_1 = hat j - 4 hat k`

` vec b_1 xx vec b_2 = | (hat i , hat j , hat k) , ( -1 , 1, - 2) ,( 1 ,2 , -2) | = 2 hat i - 4 hat j - 3 hat k`

` | vec b_1 xx vec b_2 | = sqrt ( 4 + 16 + 9) = sqrt (29)` and

` ( vec a_2 - vec a_1 ) . ( vec b_1 xx vec b_2) = ( hat j - 4 hat k) . ( 2 hat i - 4 hat j - 3 hat k) = -4 + 12 = 8`

The shortest distance between two lines is

` d = | (( vec a_2 - vec a_1 ) . ( vec b_1 xx vec b_2))/ ( | vec b_1 xx vec b_2 | ) |= | 8/sqrt(29) | = 8/sqrt(29)`
Q 2151745624

Find the shortest distance between the lines

` (x + 1)/7 = ( y + 1)/(-6) = (z + 1)/1 ` and ` ( x - 3)/1 = (y - 5)/(-2) = ( z - 7)/1`
Class 12 Exercise Q.No. 0
Solution:

Shortest distance between the lines

` (x - x_1)/a_1 = (y - y_1)/b_1 = ( z - z_1)/c_1 ` and

` (x - x_2)/a_2 = (y - y_2)/b_2 = ( z - z_2)/c_2 `

` ( | ( x_2 - x_1 , y_2 - y_1 , z_2 - z_1) ,( a_1 , b_1 , c_1 ),(a_2 , b_2 , c_2) | )/sqrt( (b_1c_2 - b_2c_1 )^2 + ( c_1a_2 - c_2a_1)^2 + ( a_1b_2 - a_2 b_1)^2 )`

The given lines are ` (x + 1)/7 = (y + 1)/(-6) = (z + 1)/1` and

` (x - 3)/1 = ( y - 5)/(-2) = (z - 7)/1`

These lines pass through the points `( -1, -1, -1)`

and `(3, 5, 7)`

`x_2 - x_1 = 3 - (-1) = 4, y_2 - y_1 = 5 - (-1) = 6`,

`z_2 - z_1 = 7 - (-1) = 8`

Now Numerator `N`

` = | ( x_2 - x_1 , y_2 - y_1 , z_2 - z_1) ,( a_1 , b_1 , c_1 ),(a_2 , b_2 , c_2) | = | ( 4 , 6,0),(7 , -6 ,1),( 1 ,-2 ,1) |`

` ∵ a_1 , b_1 , c_1 ` are `7, -6, 1 a_2, b_2, c_2` are `1, -2, 1`

`:. N = 4 (-6 +2) + 6 (1 - 7) + 8 (-14 + 6) = -116`

`D^r = sqrt( (b_1c_2 - b_2c_1 )^2 + ( c_1a_2 - c_2a_1)^2 + ( a_1b_2 - a_2 b_1)^2 )`

` :. D^r = sqrt( ( - 6 + 2)^2 + ( 1 - 7)^2 + ( - 14 + 6)^2)`

` = sqrt (116)`

`:. S.D . = | (-116)/sqrt(116) | = sqrt(116) = 2sqrt(29)`

Sphere

Q 2147234183

A plane P passes through the line of intersection of the planes
`2x- y + 3z = 2 , x + y - z = 1` and the point `(1, 0, 1).`

If the plane `P` touches the sphere `x^( 2) + y^(2) + z^(2) = r^( 2)` ,then
what is requal to?
NDA Paper 1 2016
(A)

` 2/sqrt(29)`

(B)

` 4/sqrt(29)`

(C)

` 5/sqrt(29)`

(D)

`1`

Solution:

The equation of plane `P` is

`(2x - y + 3z - 2) + lamda (x + y - z - 1) = 0`

It passes through `(1, 0, 1)`.

We get, `(2- 0 + 3 -2 ) lamda (1 + 0 -1 - 1) = 0`

`=> 3 - lamda =0 => lamda = 3`

Hence, equation of plane `P` is

`(2x- y + 3z- 2) + 3(x + y- z -1) = 0`

`=> 5x + 2y - 5 = 0`

Plane P touches the sphere

`x^(2) + y^(2) + z^(2) = r^(2)`

Here, the centre of sphere is `(0, 0, 0)`,

`:.` Radius of sphere, `r` = Perpendicular distance from

centre of sphere to the tangent plane.

` r = |(0+0+0-5)/sqrt(5^(2) +2^(2)) | =| (-5)/sqrt(29) |`

`=> r = 5/sqrt(29)` units
Correct Answer is `=>` (C) ` 5/sqrt(29)`
Q 2147234183

A plane P passes through the line of intersection of the planes
`2x- y + 3z = 2 , x + y - z = 1` and the point `(1, 0, 1).`

If the plane `P` touches the sphere `x^( 2) + y^(2) + z^(2) = r^( 2)` ,then
what is requal to?
NDA Paper 1 2016
(A)

` 2/sqrt(29)`

(B)

` 4/sqrt(29)`

(C)

` 5/sqrt(29)`

(D)

`1`

Solution:

The equation of plane `P` is

`(2x - y + 3z - 2) + lamda (x + y - z - 1) = 0`

It passes through `(1, 0, 1)`.

We get, `(2- 0 + 3 -2 ) lamda (1 + 0 -1 - 1) = 0`

`=> 3 - lamda =0 => lamda = 3`

Hence, equation of plane `P` is

`(2x- y + 3z- 2) + 3(x + y- z -1) = 0`

`=> 5x + 2y - 5 = 0`

Plane P touches the sphere

`x^(2) + y^(2) + z^(2) = r^(2)`

Here, the centre of sphere is `(0, 0, 0)`,

`:.` Radius of sphere, `r` = Perpendicular distance from

centre of sphere to the tangent plane.

` r = |(0+0+0-5)/sqrt(5^(2) +2^(2)) | =| (-5)/sqrt(29) |`

`=> r = 5/sqrt(29)` units
Correct Answer is `=>` (C) ` 5/sqrt(29)`
Q 2241634523

The radius of the sphere
`3x ^2 +3y^2 +3z^2 -8x+4y+8z -15= 0` is
NDA Paper 1 2015
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Given, `3x^2 + 3y^2 + 3z^2 - 8x + 4y + 8z -15 = 0`

`=> x^ 2 + y ^ 2 +z^ 2 - 8/3 x + 4/3 y +8/3 z -5=0`

Compare it with equation of a sphere

`x^2 + y^ 2 + z^2 + 2ux + 2vy + 2wz + d = 0`, we get

`2u = - 8/3 , 2v = 4/3 , 2w = 8/3 d = - 5`

`:. u =- 4/3 , v =2/3 , w =4/3 , d=-5`

Now, radius of a sphere `= sqrt ( u^2 + v^2 + w^2 - d)`

` = sqrt((16)/9 + 4/9 + (16)/9 + 5 ) = sqrt( 4 +5) = sqrt(9) = 3`
Correct Answer is `=>` (B) `3`
Q 2319680510

Curve of intersection of two spheres is
NDA Paper 1 2008
(A)

an ellipse

(B)

a circle

(C)

a parabola

(D)

None of these

Solution:

Intersection of two spheres form a circle.
Correct Answer is `=>` (B) a circle
Q 1689491317

Consider a sphere passing through the origin and the
points `(2, 1, -1), (1, 5,- 4), ( -2, 4, -6)`.

What is the radius of the sphere?
NDA Paper 1 2014
(A)

`(-1, 2,- 3)`

(B)

`(1,- 2, 3)`

(C)

`(1,2,-3)`

(D)

`(-1,-2,-3)`

Solution:

Equation of sphere passing through origin is

`x^ 2 + y^2 + z^2 + 2ux + 2vy + 2wz = 0`

which passes through the points. `(2, 1, -1), (1, 5, - 4)` and
`(-2, 4, -6)`.

`:. 4u + 2v - 2w = - 6` .......(1)

` 2 u + 10v - 8w = - 42` ..........(2)

and ` -4u + 8v - 12w = -56` ...........(3)

On solving above equations, we get

` u = 1, v = -2` and `w = 3`

Radius of sphere `= sqrt(u^2 + v^2 + w^2)`

` = sqrt( 1+4+9)`

`= sqrt(14)`
Correct Answer is `=>` (A) `(-1, 2,- 3)`
Q 2349212113

What is the equation of the sphere with unit
radius having centre at the origin?
NDA Paper 1 2013
(A)

`x^2 + y^2 + z^2 = 0`

(B)

`x^2 + y^2 + z^2 = 1`

(C)

`x^2 + y^2 + z^2 = 2`

(D)

`x^2 + y^2 + z^2 = 3`

Solution:

We know that, the standard equation of the sphere,
whose centre at `(x_1, y_1, z_1)` and radius is `r`, is

`(x- x_1 )^2 + (y- y_1)^2 + (z-z_1)^2 = r^2` ... (i)

Given that,

Centre `-> (0, 0, 0)` and radius `-> 1`

So, the required equation of sphere is

`(x- 0)^2 + (y- 0)^2 + (z- 0)^2 = (1)^2`

`=> x^2 + y^2 +z^2=1`
Correct Answer is `=>` (B) `x^2 + y^2 + z^2 = 1`
Q 2339745612

What is the diameter of the i sphere
`x^2 +y^2 +z^2 -4x+6y-8z-7=0`?
NDA Paper 1 2012
(A)

4 units

(B)

5 units

(C)

6 units

(D)

12 units

Solution:

The equation of sphere,

`x^2+ y^2 + z^2 - 4x+ 6y- Bz-7= 0`

Compare with,

`ax^2 + by^2 + cz^2 + 2ux + 2vy + 2wz + d = 0`

Here, `u = -2, v = 3, w = - 4` and `d = - 7`

`:.` Radius of sphere `= sqrt (u^2 +v^2 +w^2 -d)`

`= sqrt ( (-2)^2 + (3)^2 + (-4)^2 +7)`

`= sqrt (4+9 +16 +7) = sqrt (36) = 6`

`:.` Diameter `= 2 xx` Radius `= 2 xx 6 = 12` units
Correct Answer is `=>` (D) 12 units
Q 2349145913

What is the locus of points of intersection of a
sphere and a plane?
NDA Paper 1 2011
(A)

Circle

(B)

Ellipse

(C)

Parabola

(D)

Hyperbola

Solution:

If a sphere is intersected by a plane, then the locus of
set of points common to both sphere and plane is always a circle.
Correct Answer is `=>` (A) Circle
Q 2359156014

If the equation to sphere passing through origin
and the points `(-1, 0, 0), (0,- 2, 0)` and `(0, 0,-3)` is
`x^2 + y^2 + z^2 + f(x, y, z) = 0`, then what is
`f(x,y,z)` equal to?
NDA Paper 1 2011
(A)

`-x -2y- 3z`

(B)

`x + 2y + 3z`

(C)

`x + 2y + 3z- 1`

(D)

`x+2y+3z+1`

Solution:

The general equation of sphere is

`x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0` ... (i)

which passes through the origin and the points `( -1, 0, 0)`,

`(0, -2, 0)` and `(0, 0,- 3)`, then

`0+0+0+0+0+0+d=0`

`=> d = 0`

and `1 + 0 + 0 - 2u + 0 + 0 + d = 0`

`=> 2u = d + 1`

`=> 2u= 0+1 => u =1/2`

and `0 + 4 + 0 + 0 - 4v + 0 + d = 0`

` => 4v = 4 + d`

`=> 4v = 4 + 0`

`=> v = 1`

and `0 + 0 + 9 + 0 + 0 - 6w + d = 0`

`=> 6w=9+d`

`=> 6 w = 9 + 0`


` :. w= 3/2`

On putting these values in Eq. (i), we get

`x^2 + y^2 + z^2 + 2 (1/2) x + 2(1) y +2 (3/2) z + 0 =0`

`=> x^2 +y^2 +z^2 +x +2y +3z =0`

Comparing it with,

`x^2 + y^2 + z^2 + f(x, y, z) = 0`

`=> f(x, y, z) = x + 2y + 3z`
Correct Answer is `=>` (B) `x + 2y + 3z`
Q 2456834774

The radius of circle made by meeting the plane
`x + 2y + 2z + 7 = 0` to the sphere
`x^2 + y ^2 + z ^2 + 2x- 2y - 4z -19 = 0` is
UPSEE 2010
(A)

`5`

(B)

`4`

(C)

`3`

(D)

None of these

Solution:

The centre of sphere `C = ( -1, 1, 2)` and
radius of sphere `r = sqrt(1+ 1 + 4 + 9) = 5`
Now, perpendicular distance of centre from
plane is

`p=(-1+2+4+7)/(sqrt (1+4+4))=4`

`:.` Radius of circle `r' = sqrt(r^2-p^2) =sqrt(25-16)`

`=3`
Correct Answer is `=>` (C) `3`
Q 2379023816

What should be the value of `k` for which the
equation `3x^2 + 3y^2 + (k + 1) z^2 + x-y +z =0`
represents the sphere?
NDA Paper 1 2013
(A)

`3`

(B)

`2`

(C)

`1`

(D)

`-1`

Solution:

Given equation of conic

`3x^2+ 3y2+ (k+ 1)z^2 + x- y+ z=0`

It this conic represent a sphere, then

Coefficient of `x^2 =` Coefficient of `y^2 =` Coefficient of `z^2`

`3=3=k+1`

`:. k=3-1=2`
Correct Answer is `=>` (B) `2`
Q 2480801717

What is the centre of the sphere
`ax^2 + by^2 + cz^2 - 6x = 0`, if the radius is `1` unit ?

NDA Paper 1 2007
(A)

`(0, 0, 0)`

(B)

` (1 0, 0)`

(C)

`(3, 0, 0)`

(D)

Cannot be determined

Solution:

For the sphere,

Coefficient of `x =` Coefficient of `y =` Coefficient of `z`

`=> a=b=c`

So, `ax^2 + by^2 + cz^2 - 6x = 0`

`=> x^2 + y^2 + z^2 - (6x)/a = 0`

`:.` Centre `= (3/a ,0, 0)`

Given that, radius `= 1`

`=> sqrt ((3/a)^2 +0 +0 ) =1 => 3/a =1`

`:. a= 3`

`:.` Centre `= (1, 0, 0)`
Correct Answer is `=>` (B) ` (1 0, 0)`
Q 2480801717

What is the centre of the sphere
`ax^2 + by^2 + cz^2 - 6x = 0`, if the radius is `1` unit ?

NDA Paper 1 2007
(A)

`(0, 0, 0)`

(B)

` (1 0, 0)`

(C)

`(3, 0, 0)`

(D)

Cannot be determined

Solution:

For the sphere,

Coefficient of `x =` Coefficient of `y =` Coefficient of `z`

`=> a=b=c`

So, `ax^2 + by^2 + cz^2 - 6x = 0`

`=> x^2 + y^2 + z^2 - (6x)/a = 0`

`:.` Centre `= (3/a ,0, 0)`

Given that, radius `= 1`

`=> sqrt ((3/a)^2 +0 +0 ) =1 => 3/a =1`

`:. a= 3`

`:.` Centre `= (1, 0, 0)`
Correct Answer is `=>` (B) ` (1 0, 0)`
Q 2886012877

What is the equation of the sphere which has its
centre at `(6,- 1, 2)` and touches the plane
`2x - y + 2z - 2 = 0` ?

(A)

`x^2 + y^2 + z^2 + 12x - 2y+ 4z+ 16= 0`

(B)

`x^2 + y^2 + z^2 + 12x-2y+ 4z-16= 0`

(C)

`x^2 + y^2 + z^2 -12x+ 2y- 4z+ 16= 0`

(D)

`x^2 + y^2 + z^2 -12x+ 2y- 4z+ 25= 0`

Solution:

Given centre of sphere is `(6, -1 ,2 )`

`:.` Radius `= (2 (6) - 1 (-1) +2 (2) -2 )/(sqrt (4+1+4) ) =15/3 = 5`

`:.` Equation of sphere is ` ( x-6)^2 + (y+1)^2 + (z-2)^2 = 5^2`

`=> x^2 + y^2 + z^2 -12 x + 2y -4z +16 = 0`
Correct Answer is `=>` (C) `x^2 + y^2 + z^2 -12x+ 2y- 4z+ 16= 0`
Q 2420101011

The equation of a sphere is
`x^2 + y^2 + z^2 - 10 z = 0`. If one end point of a
diameter of the sphere is `(-3, -4, 5)`, then what is
the other end point?
NDA Paper 1 2007
(A)

`(-3,- 4,- 5)`

(B)

`(3, 4, 5)`

(C)

`(3, 4, - 5)`

(D)

`(-3, 4,- 5)`

Solution:

The equation of sphere is

`x^2 + y^2 + z^2 - 10z = 0`

So, the centre of sphere is `C(0, 0, 5)`.

Coordinates of one end point of a diameter of the sphere is

`A(-3,- 4, 5)`.

Let coordinates of another end point of this diameter is

`B(x_1, y_1, z_1)`.

`:. (-3 +x_1)/2 =0 => x_1 =3`

`(-4+y_1)/2 =0 => y_1= 4`

and `(5+z_1)/2 =5 => z_1 =5`

So, the required coordinates are `(3, 4, 5)`.
Correct Answer is `=>` (B) `(3, 4, 5)`
Q 1686280177

The radius of the sphere

`3x ^2 +3y^2 +3z^2 -8x+4y+8z -15= 0` is
DSSB Paper 1 2015
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Given, `3x^2 + 3y^2 + 3z^2 - 8x + 4y + 8z -15 = 0`

`=> x^ 2 + y ^ 2 +z^ 2 - 8/3 x + 4/3 y +8/3 z -5=0`

Compare it with equation of a sphere

`x^2 + y^ 2 + z^2 + 2ux + 2vy + 2wz + d = 0`, we get

`2u = - 8/3 , 2v = 4/3 , 2w = 8/3 d = - 5`

`:. u =- 4/3 , v =2/3 , w =4/3 , d=-5`

Now, radius of a sphere `= sqrt ( u^2 + v^2 + w^2 - d)`

` = sqrt((16)/9 + 4/9 + (16)/9 + 5 ) = sqrt( 4 +5) = sqrt(9) = 3`
Correct Answer is `=>` (B) `3`
Q 2319580419

If the radius of the sphere

`x^2 + y^2 +z^2 - -6x - 8y + 10z + llambda = 0`
is unity, what is the value of `lambda` ?
NDA Paper 1 2008
(A)

`49`

(B)

`7`

(C)

`-49`

(D)

`-7`

Solution:

Radius of sphere

`= sqrt ( (-6/2)^2 + (-8/2)^2 + (10/2)^2 -lambda)`

`=> 1= 9+16 +25 - lambda` (given)

`=> lambda = 49`
Correct Answer is `=>` (A) `49`
Q 2349378213

Find the coordinate of the point, which is
equidistant from the points `(0, 0 0) , (2,0,0),
(0, 4, 0)`, and `(0, 0, 6)`, respectively
NDA Paper 1 2008
(A)

`(1, 2, 3)`

(B)

`(2, 3, 1)`

(C)

`(3, 1, 2)`

(D)

`(1,3,2)`

Solution:

Let the equation of sphere which passes through the
given four points be


`x^2 + y^2 + z^2 + 2ux + 2vy + 2wz +d =0`

`:. 0+0+0+0+0+0 + d =0`

`=> d =0` ..........(i)

So, `(2)^2 + 0 + 0 + 4u + 0+0 +d =0`

`4u + d = -4`

`=> 4u + 0 = -4 => u = -1` .......................(ii)

Therefore,

`0 + (4)^2 + 0+ 0 + 8v + 0 +d =0`

`=> 16 + 8v + d =0`

`=> 8v + 0 = -16 => v =-2` ...........(iii)

Now, `0+0 + (6)^2 + 0+ 0 + 12w +d =0`

`=> 36 +12w +d =0`

`=> 12w + 0 = -36 => w = -3`.................(iv)

`:.` Centre of sphere `= (1, 2, 3)`

So, the required point is the centre of sphere, which is equidistant
from given four points.
Correct Answer is `=>` (A) `(1, 2, 3)`
Q 2389278117

Under what condition does the equation
`x^2 + y^2 + z^2 + 2ux + 21 vy + 2wz + d = 0` represent
a real sphere'?
NDA Paper 1 2009
(A)

`u^2 + v^2 + w^2 = d^2`

(B)

`u^2 + v^2 + w^2 > d`

(C)

`u^2 + v^2 + w^2 < d`

(D)

`u^2 + v^2 + w^2 < d^2`

Solution:

The given equation represents a real sphere , if

`u^2 + v^2 + w^2 > d` (by definition)
Correct Answer is `=>` (B) `u^2 + v^2 + w^2 > d`
Q 2379178016

What is the equation of the sphere which has its
centre at `(6, -1, 2)` and touches the plane
`2x - y + 2z - 2 = 0` ?
NDA Paper 1 2009
(A)

`x^2 + y^2 + z^2 + 12x- 2y + 4z + 16 =0`

(B)

`x^2 + y^2 + z^2 + 12x - 2 y + 4z - 16 = 0`

(C)

`x^2 + y^2 + z^2 - 12x + 2y- 4z + 16 = 0`

(D)

`x^2 + y^2 + z^2 - 12x + 2y- 4z + 25 = 0`

Solution:

Given the centre of sphere to be `(6, -1, 2)`.

`:.` Radius = Perpendicular distance to the plane from the centre

`:.` Radius `= | (2(6) -1 (-1) +2(2) -2)/(sqrt (4+1+4) ) | = 15/3 =5`

`:.` Equation of sphere is

`(x-6)^2 + (y+1)^2 + (z-2)^2 =5^2`

`=> x^2 + y^2 + z^2 - 12x + 2y- 4z + 16 = 0`
Correct Answer is `=>` (C) `x^2 + y^2 + z^2 - 12x + 2y- 4z + 16 = 0`
Q 2339056812

What is the radius of the sphere

`x^2 +y^2 + z^2 -x -y -z=0`?
NDA Paper 1 2011
(A)

`sqrt (3/4)`

(B)

`sqrt (1/2)`

(C)

`sqrt (3/2)`

(D)

`1/3`

Solution:

The given equation of sphere is

`x^2 + y^2 + z^2 - x - y- z = 0`


Compare with `x^2 + y^2 + z^2 + 2ux + 2vy + 2k + d = 0`

`=> u = -1/2 , v = -1/2 , w=-1/2` and `d=0`

`:.` Radius of sphere `= sqrt (u^2 + v^2 + w^2 -d)`

`= sqrt (1/4 +1/4 +1/4 -0) = sqrt (3/4)`
Correct Answer is `=>` (A) `sqrt (3/4)`
Q 2336167972

If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, then what is `(x + y + z)^3`
equal to?
NDA Paper 1 2007
(A)

`1`

(B)

`3`

(C)

`3xyz`

(D)

`27xyz`

Solution:

`x^(1/3) + y^(1/3) + z^(1/3) = 0`

`=> x^(1/3)+y^(1/3) = - z^(1/3) ` .....(i)


`=> (x^(1/3)+y^(-1/3))^3 = -z`

`=> x+y+3x^(1/3)y^(1/3)(x^(1/3)+y^(1/3)) = -z`


`=> x+y+z = 3x^(1/3)y^(1/3)z^(1/3)` from eq (i)

`therefore (x+y+z)^3 = (3x^(1/3)y^(1/3)z^(1/3))^3`

`= 27xyz`
Correct Answer is `=>` (D) `27xyz`
 
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