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Must Do Problems Of Three Dimensional Geometry For NDA

Must Do Problems Of Three Dimensional Geometry

Must Do Problems Of Three Dimensional Geometry
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 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 2876312276

The angle between the lines with direction ratios
`(1, 0, pm cos alpha )` is `60^o`. What is the value of `alpha` ?

(A)

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

(B)

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

(C)

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

(D)

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

Solution:

`cos 60^o = ( [ 1 xx 1 +0 xx 0 + ( cos alpha ) (-cos alpha ) ] )/( [ (sqrt (1^2 + (0)^2 + cos^2 alpha ) )/( 1^2 + (0)^2 + (- cos alpha) ) *2 ] )`

`=> 1/2 = (1- cos^2 alpha )/( sqrt (1+ cos62 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 )`

[applying componcndo and dividendo ]

`=> 3/(-1) = 1/(- cos^2 alpha ) => cos alpha = 1/sqrt3`

`:. alpha = cos^(-1) (1/sqrt 3)`
Correct Answer is `=>` (B) `cos^(-1) (1/sqrt 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 1626380271

The direction ratios of the line perpendicular to the lines

with direction ratios `< 1, - 2, - 2 >` and `< 0, 2, 1 >` are
DSSB 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 1563234145

The two lines `x = my + n, z = py + q` and
`x = m'y + n', z = p' y + q'` are perpendicular
to each other, if
VIT 2015
(A)

`mm' + pp' =1`

(B)

`m/m' + p/p' =-1`

(C)

`m/m' + p/p' =1`

(D)

`mm' + pp' =-1`

Solution:

Given equations of lines are

`x = my + n, z = py + q`

and `x = m' y + n', z = p' y + q'`

These equations can be rewritten as

` (x-n)/m = (y-0)/1 = (z-q)/(p)`

` (x-n')/m' = (y-0)/1 = (z-q')/(p')`

These lines will be perpendicular, if

`mm' + 1 + pp' = 0`

`=>mm' + pp' =-1`
Correct Answer is `=>` (D) `mm' + pp' =-1`
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 2856112074

The equation of line through the point (1, 2, 3)
para e to line `(x-4)/2 = (y+1)/( -3) = (z+10)/8` is

(A)

`(x-1)/2 = ( y-2)/(-3) = (z-3)/8`

(B)

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

(C)

`(x-4)/1 = (y+1)/2 = (z+10)/3`

(D)

None of these

Solution:

Let the equation of line passing
through (1, 2, 3) is

`(x-1)/a = (y-2)/b = (z-3)/c`

But it is parallel to the given line

`:. a/2 = b/(-3) = c/8`

Hence, the required line is

`(x-1)/2 = (y-2)/(-3) = (z-3)/8`
Correct Answer is `=>` (A) `(x-1)/2 = ( y-2)/(-3) = (z-3)/8`
Q 2435334262

The angle between two diagonals of a cube will
be
UPSEE 2012
(A)

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

(B)

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

(C)

variable

(D)

None of these

Solution:

Let the cube be of side `a`.

`O(0, 0, 0), D(a ,a,a ), B(0, a, 0), G(a, 0, a)`,

then equation of `OD` and `BG` are `x/a = y/a =z/a` and


`x/a = (y-a)/(-a) = z/a`, respectively.

Hence, angle betwc:en `OD` and `BG` is

`cos^(-1) ((a^2 - a^2 + a^2)/(sqrt (3a^2) * sqrt (3a^2))) = cos^(-1) (1/3)`

Note -- Students should remember this question as a fact.
Correct Answer is `=>` (B) `cos^(-1) (1/3)`
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 1563080845

The vector equation `r=i-2j-k+t(6j-k)` represents a straight line passing through the points:
BITSAT 2012
(A)

`(0, 6, -1)` and `(1, -2, -1)`

(B)

`(0, 6, -1) ` and `(-1, -4, -2)`

(C)

`(1, -2, -1) ` and `(1, 4, -2)`

(D)

`(1, -2, -1) ` and `(0, -6, 1)`

Solution:

Cartesian representation of the given line is,

`\ frac{x-1}{0}=\ frac{y+2}{6}=\ frac{z+1}{-1}=t`

So any point on the given line is of the form `(1,6t-2, -t-1)`

where `t` can be any real numbers

so for `t=0` and `1` the corresponding points are `(1,-2,-1)` and `(1,4,-2)`

You can check other options does not satisfy above point for any `t`.
Correct Answer is `=>` (C) `(1, -2, -1) ` and `(1, 4, -2)`
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 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 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 1561334225

The point of intersection of the line `(x-1)/3 = (y+2)/4 = (z-3)/(-2)` and plane `2x-y + 3z -1 =0` is:
BITSAT 2005
(A)

`(10, - 10, 3)`

(B)

`(10, 10, - 3)`

(C)

`(-10, 10, 3)`

(D)

None of these

Solution:

Given equation of line is

`(x-1)/3 = (y+2)/4 = (z-3)/(-2) = k` say

Any point on the line is

`(3k + 1, 4k- 4- 2k + 3)`

If the given line intersect the plane

`2.x - y + 3z -1 = 0`, then any point on the line lies in the plane.

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

` => -4k +12=0 = k=3`

`:.` Point is `(9 + 1, 12 - 2, - 6 + 3)`

i.e., `(10, 10, -3)`.
Correct Answer is `=>` (B) `(10, 10, - 3)`
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)`
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 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)`
 
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