`color{red} ♦` Plane

`color{red} ♦` Equation of a plane in normal form

`color{red} ♦` Equation of a plane perpendicular to a given vector and passing through a given point

`color{red} ♦` Equation of a plane passing through three non collinear points

`color{red} ♦` Intercept form of the equation of a plane

`color{red} ♦` Plane passing through the intersection of two given planes

`color{red} ♦` Coplanarity of Two Lines

`color{red} ♦` Angle between Two Planes

`color{red} ♦` Distance of a Point from a Plane

`color{red} ♦` Angle between a Line and a Plane

`color{red} ♦` Equation of a plane in normal form

`color{red} ♦` Equation of a plane perpendicular to a given vector and passing through a given point

`color{red} ♦` Equation of a plane passing through three non collinear points

`color{red} ♦` Intercept form of the equation of a plane

`color{red} ♦` Plane passing through the intersection of two given planes

`color{red} ♦` Coplanarity of Two Lines

`color{red} ♦` Angle between Two Planes

`color{red} ♦` Distance of a Point from a Plane

`color{red} ♦` Angle between a Line and a Plane

● A plane is determined uniquely if any one of the following is known :

(i) the normal to the plane and its distance from the origin is given, i.e., equation of a plane in normal form.

(ii) it passes through a point and is perpendicular to a given direction.

(iii) it passes through three given non collinear points.

Now we shall find vector and Cartesian equations of the planes.

(i) the normal to the plane and its distance from the origin is given, i.e., equation of a plane in normal form.

(ii) it passes through a point and is perpendicular to a given direction.

(iii) it passes through three given non collinear points.

Now we shall find vector and Cartesian equations of the planes.

`=>` Consider a plane whose perpendicular distance from the origin is `d (d ≠ 0).` Fig.

`=>` If `bar (ON)` is the normal from the origin to the plane, and `hat n` is the unit normal vector along `bar (ON)` . Then `bar (ON) = d hat n`.

`=>` Let P be any point on the plane. Therefore,

`bar (NP)` is perpendicular to `bar (ON),` Therefore, `bar (ON) * bar (ON) = 0` ..........(1)

`=>` Let `vec r` be the position vector of the point P, then `bar (NP) = vec r -d hat n` (as `bar (ON) + bar (NP) = bar (OP)`)

Therefore, (1) becomes

` ( vec r - d hat n) * d hat n = 0`

or `( vec r - d hat n) * hat n =0` `( d ≠ 0)`

or `vec r * hat n - d hat n * hat n = 0`

i.e., `color{red}{vec r * hat n = d}` (as `hat n * hat n =1`) ..........(2)

This is the vector form of the equation of the plane.

`color{brown} "Cartesian form of the equation of the plane "`

`=>` Equation (2) gives the vector equation of a plane, where `hat n` is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane. Then

`bar (OP) = vec r = x hat i + y hat j + z hat k`

`=>` Let `l, m, n` be the direction cosines of `hat n` . Then

`hat n = l hat i + m hat j + n hat k`

Therefore, (2) gives

`( x hat i + y hat j + z hat k) * (l hat i + m hat j + n hat k) =d`

i.e., `color{red}{lx+ my +nz = d}` .............(3)

`=>` This is the cartesian equation of the plane in the normal form.

`=>` If `bar (ON)` is the normal from the origin to the plane, and `hat n` is the unit normal vector along `bar (ON)` . Then `bar (ON) = d hat n`.

`=>` Let P be any point on the plane. Therefore,

`bar (NP)` is perpendicular to `bar (ON),` Therefore, `bar (ON) * bar (ON) = 0` ..........(1)

`=>` Let `vec r` be the position vector of the point P, then `bar (NP) = vec r -d hat n` (as `bar (ON) + bar (NP) = bar (OP)`)

Therefore, (1) becomes

` ( vec r - d hat n) * d hat n = 0`

or `( vec r - d hat n) * hat n =0` `( d ≠ 0)`

or `vec r * hat n - d hat n * hat n = 0`

i.e., `color{red}{vec r * hat n = d}` (as `hat n * hat n =1`) ..........(2)

This is the vector form of the equation of the plane.

`color{brown} "Cartesian form of the equation of the plane "`

`=>` Equation (2) gives the vector equation of a plane, where `hat n` is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane. Then

`bar (OP) = vec r = x hat i + y hat j + z hat k`

`=>` Let `l, m, n` be the direction cosines of `hat n` . Then

`hat n = l hat i + m hat j + n hat k`

Therefore, (2) gives

`( x hat i + y hat j + z hat k) * (l hat i + m hat j + n hat k) =d`

i.e., `color{red}{lx+ my +nz = d}` .............(3)

`=>` This is the cartesian equation of the plane in the normal form.

`color{red} "Key Point "` - Equation (3) shows that if `vec r * (a hat i + b hat j + c hat k) = d` is the vector equation of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b and c are the direction ratios of the normal to the plane.

Q 3107678588

Find the vector equation of the plane which is at a distance of `6/(sqrt 29)`

from the origin and its normal vector from the origin is `2 hat i -3 hat j + 4 hat k` .Also find its cartesian form.

Class 12 Chapter 11 Example 13

from the origin and its normal vector from the origin is `2 hat i -3 hat j + 4 hat k` .Also find its cartesian form.

Class 12 Chapter 11 Example 13

Let `vec n = 2 hat i -3 hat j + 4 hat k` . Then

`hat n = (vec n )/(| vec n | ) = (2 hat i -3 hat j+4 hat k)/(sqrt ( 4+9 +16) ) = (2 hat i -3hat j + 4 hat k)/(sqrt 29)`

Hence, the required equation of the plane is

`vec r * (2/(sqrt29 ) hat i + (-3)/(sqrt 29) hatj + 4/(sqrt29) hat k) = 6/(sqrt 29)`

Q 3117678589

Find the direction cosines of the unit vector perpendicular to the plane

`vec r * (6 hat i -3 hat j -2 hat k) +1 = 0` passing through the origin.

Class 12 Chapter 11 Example 14

`vec r * (6 hat i -3 hat j -2 hat k) +1 = 0` passing through the origin.

Class 12 Chapter 11 Example 14

The given equation can be written as

`vec r * ( -6 hat i + 3 hat j + 2 hat k ) =1` ......(1)

Now ` | - 6 hat i +3 hat j + 2 hat k | = sqrt(36 + 9 +4) = 7`

Therefore, dividing both sides of (1) by 7, we get

`vec r * ( -6/7 hat i + 3/7 hat j + 2/7 hat k) = 1/7`

which is the equation of the plane in the form `vec r * hat n = d` .

This shows that `hat n = -6/7 hat i + 3/7 hat j + 2/7 hat k` is a unit vector perpendicular to the

plane through the origin. Hence, the direction cosines of `hat n` are ` (-6)/7 , 3/7 , 2/7`

Q 3117778680

Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.

Class 12 Chapter 11 Example 15

Class 12 Chapter 11 Example 15

Since the direction ratios of the normal to the plane are 2, –3, 4; the direction

cosines of it are

`2/(sqrt (2^2 + (-3)^2 + 462) ) , (-3)/(sqrt (2^2 + (-3)^2 +4^2) ) , 4/( sqrt (2^2+ (-3)^2 + 4^2) )` , i.e., ` 2/(sqrt 29) , (-3)/( sqrt 29) , 4/(sqrt 29)`

Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i.e., 2x – 3y + 4z = 6 throughout by

`sqrt29` , we get

`2/(sqrt29 ) x + (-3)/(sqrt29) y + 4/(sqrt29) z = 6/(sqrt 29)`

This is of the form lx + my + nz = d, where d is the distance of the plane from the

origin. So, the distance of the plane from the origin is `6/(sqrt 29)` .

Q 3127778681

Find the coordinates of the foot of the perpendicular drawn from the

origin to the plane 2x – 3y + 4z – 6 = 0.

Class 12 Chapter 11 Example 16

origin to the plane 2x – 3y + 4z – 6 = 0.

Class 12 Chapter 11 Example 16

Let the coordinates of the foot of the perpendicular P from the origin to the

plane is `(x_1, y_1, z_1)` (Fig 11.11).

Then, the direction ratios of the line OP are

`x_1, y_1, z_1` .

Writing the equation of the plane in the normal

form, we have

`2/(sqrt29) x - 3/(sqrt29 ) y + 4/(sqrt29 ) z = 6/(sqrt29)`

where `2/(sqrt29 ) , (-3)/(sqrt29) , 4/(sqrt29)` are the direction cosines of the OP.

Since d.c.’s and direction ratios of a line are proportional, we have

` (x_1)/( 2/(sqrt29)) = (y_1)/( (-3)/(sqrt29) ) = (z_1)/(4/(sqr29 )) = k`

i.e., `x_1= (2k)/(sqrt29) , y_1 (-3 k)/( sqrt29) , z_1 = (4 k)/(sqrt29)`

Substituting these in the equation of the plane, we get k `= 6/(sqrt29 )`

Hence, the foot of the perpendicular is ` (12/29 , (-18)/29 , 24/29 )`

Note : If `d` is the distance from the origin and `l, m, n` are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is `(ld, md, nd).`

`=>` In the space, there can be many planes that are perpendicular to the given vector, but through a given point `P(x_1, y_1, z_1)`, only one such plane exists (see Fig).

`=>` Let A plane pass through a point A with position vector`vec a` and perpendicular to the vector`bar N`

Let `vec r` be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).

`=>` Then the point P lies in the plane if and only if

`bar (AP) ` is perpendicular to `bar (N)` , i.e., `bar (AP) *bar N =0`. But

`bar (AP) = vec r - vec a` . Therefore `(vec r - vec a) * vec N = 0` .......(1)

This is the vector equation of the plane.

`color{brown} "Cartesian form"`

`=>` Let the given point A be `(x_1, y_1, z_1)`, P be (x, y, z) and direction ratios of `bar N` are A, B and C. Then,

`vec a = x_1 hat i + y_1 hat j + z_1 hat k , vec r = x hat i + y hat j + z hat k` and `bar(N) = A hat i + B hat j + C hat k`

Now `(vec r - vec a) * vec N= 0`

So ` [ (x-x_1) hat i + (y-y_1) hat j + (z-z_1) hat k]* ( A hat i +B hat j + C hat k) =0`

i.e., `color{red}{A (x- x_1) +B(y- y_1) +C(z-z_1) = 0}`

`=>` Let A plane pass through a point A with position vector`vec a` and perpendicular to the vector`bar N`

Let `vec r` be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).

`=>` Then the point P lies in the plane if and only if

`bar (AP) ` is perpendicular to `bar (N)` , i.e., `bar (AP) *bar N =0`. But

`bar (AP) = vec r - vec a` . Therefore `(vec r - vec a) * vec N = 0` .......(1)

This is the vector equation of the plane.

`color{brown} "Cartesian form"`

`=>` Let the given point A be `(x_1, y_1, z_1)`, P be (x, y, z) and direction ratios of `bar N` are A, B and C. Then,

`vec a = x_1 hat i + y_1 hat j + z_1 hat k , vec r = x hat i + y hat j + z hat k` and `bar(N) = A hat i + B hat j + C hat k`

Now `(vec r - vec a) * vec N= 0`

So ` [ (x-x_1) hat i + (y-y_1) hat j + (z-z_1) hat k]* ( A hat i +B hat j + C hat k) =0`

i.e., `color{red}{A (x- x_1) +B(y- y_1) +C(z-z_1) = 0}`

Q 3137778682

Find the vector and cartesian equations of the plane which passes through

the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1.

Class 12 Chapter 11 Example 17

the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1.

Class 12 Chapter 11 Example 17

We have the position vector of point (5, 2, – 4) as `vec a =5 hat i + 2 hat j −4 hat k` and the

normal vector `vec N` perpendicular to the plane as` vec N = 2 hat i +3 hat j − hat k`

Therefore, the vector equation of the plane is given by `( vec r − vec a) * vec N = 0`

or ` [ vec r - (5 hat i + 2 hat j - 4 hat k ) ] * (2 hat i +3 hat j - hat k) = 0`......(1)

Transforming (1) into Cartesian form, we have

` [ (x-5) hat i + (y-2) hat j + (z+4) hat k ] * (2 hat i +3 hat j - hat k) =0`

or 2(x −5)+3( y −2)−1(z + 4)=0

i.e. 2x + 3y – z = 20

which is the cartesian equation of the plane.

`=>` Let `R, S` and `T` be three non collinear points on the plane with position vectors `vec a ,vec b` and `vec c` respectively (Fig).

`=>` The vectors `bar (RS) ` and `vec (RT)` are in the given plane. Therefore, the vector `bar (RS) xx vec (RT)` is perpendicular to the plane containing points R, S and T. Let `vec r` be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector `bar (RS) xx bar (RT) ` is

`(vec r - vec a) * (bar (RS) xx bar (RT) ) = 0`

or `color{red}{( vec r - vec a) xx [ (vec b - vec a) xx (vec c - vec a) ] = 0}` ........(1)

`=>` This is the equation of the plane in vector form passing through three noncollinear points.

`=>` These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.

`color{brown} "Cartesian form"`

`=>` Let `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)` be the coordinates of the points R, S and T respectively. Let `(x, y, z)` be the coordinates of any point P on the plane with position vector `vec r` . Then

`bar (RP) = (x-x_1)hat i + (y-y_1) hat j + (z-z_1) hat k`

`bar (RS) = (x_2 - x_1) hat i + (y_2 - y_1) hat j + (z_2 - z_1) hat k`

`bar (RT) = (x_3 -x_1) hat i + (y_3- y_1) hat j + (z_3 -z_1) hat k`

`=>` Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have

`=>` which is the equation of the plane in Cartesian form 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)`.

`=>` The vectors `bar (RS) ` and `vec (RT)` are in the given plane. Therefore, the vector `bar (RS) xx vec (RT)` is perpendicular to the plane containing points R, S and T. Let `vec r` be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector `bar (RS) xx bar (RT) ` is

`(vec r - vec a) * (bar (RS) xx bar (RT) ) = 0`

or `color{red}{( vec r - vec a) xx [ (vec b - vec a) xx (vec c - vec a) ] = 0}` ........(1)

`=>` This is the equation of the plane in vector form passing through three noncollinear points.

`color{red} " Key Point"`- It's necessary to say that the three points had to be non collinear because If the three points were on the same line, then there will be many planes that will contain them (Fig).

`=>` These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.

`color{brown} "Cartesian form"`

`=>` Let `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)` be the coordinates of the points R, S and T respectively. Let `(x, y, z)` be the coordinates of any point P on the plane with position vector `vec r` . Then

`bar (RP) = (x-x_1)hat i + (y-y_1) hat j + (z-z_1) hat k`

`bar (RS) = (x_2 - x_1) hat i + (y_2 - y_1) hat j + (z_2 - z_1) hat k`

`bar (RT) = (x_3 -x_1) hat i + (y_3- y_1) hat j + (z_3 -z_1) hat k`

`=>` Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have

`\ "Equation of the plane" color{red} { | (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}`

`=>` which is the equation of the plane in Cartesian form 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)`.

Q 3147778683

Find the vector equations of the plane passing through the points

R(2, 5, – 3), S (– 2, – 3, 5) and T(5, 3,– 3).

Class 12 Chapter 11 Example 18

R(2, 5, – 3), S (– 2, – 3, 5) and T(5, 3,– 3).

Class 12 Chapter 11 Example 18

Let `vec a = 2 hat i + 5 hat j -3 hat k , vec b = -2 hat i -3 hat j + 5 hat k , vec c = 5 hat i + 3 hat j - 3 hat k`

Then the vector equation of the plane passing through `vec a , vec b` and `vec c`

and is given by

`(vec r - vec a) * ( bar (RS) xx bar (RT) ) = 0` (why ?)

or ` (vec r - vec a) * [ ( vec b - vec a) xx ( vec c - vec a) ] = 0`

i.e., ` [ vec r - (2 hat i + 5 hat j -3 hat k) ] * [ (-4 hat i - 8 hat j + 8 hat k ) xx (3 hat i -2 hat j) ] = 0`

`=>` In this section, we shall deduce the equation of a plane in terms of the intercepts made by the plane on the coordinate axes. Let the equation of the plane be

`Ax + By + Cz + D = 0 \ \ (D ≠ 0)` ... (1)

`=>` Let the plane make intercepts a, b, c on `x, y` and `z` axes, respectively (Fig).

`=>` Hence, the plane meets x, y and z-axes at `(a, 0, 0), (0, b, 0), (0, 0, c),` respectively.

Therefore ` Aa + D = 0` or `A = (-D)/a`

`Bb +D =0` or `B = (-D)/b`

`Cc +D =0` or `C= (-D)/c`

`=>` Substituting these values in the equation (1) of the plane and simplifying, we get

`color{red}{x/a + y/b + z/c = 1}` .......(1)

which is the required equation of the plane in the intercept form.

`Ax + By + Cz + D = 0 \ \ (D ≠ 0)` ... (1)

`=>` Let the plane make intercepts a, b, c on `x, y` and `z` axes, respectively (Fig).

`=>` Hence, the plane meets x, y and z-axes at `(a, 0, 0), (0, b, 0), (0, 0, c),` respectively.

Therefore ` Aa + D = 0` or `A = (-D)/a`

`Bb +D =0` or `B = (-D)/b`

`Cc +D =0` or `C= (-D)/c`

`=>` Substituting these values in the equation (1) of the plane and simplifying, we get

`color{red}{x/a + y/b + z/c = 1}` .......(1)

which is the required equation of the plane in the intercept form.

Q 3117178989

Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and

z-axis respectively.

Class 12 Chapter 11 Example 19

z-axis respectively.

Class 12 Chapter 11 Example 19

Let the equation of the plane be

`x/a + y/b + z/c = 1` ......(1)

Here `a=2 , b =3 , c=4`

Substituting the values of a, b and c in (1), we get the required equation of the

plane as `x/2 + y/3 + z/4 =1` or `6x + 4y +3z= 12`

`=>` Let `π_1` and `π_2` be two planes with equations `vec r * hat n_1 = d_1` and `vec r * hat n_2 = d_2` respectively. The position vector of any point on the line of intersection must satisfy both the equations (Fig).

`=>` If `vec t` is the position vector of a point on the line, then

`vec t * hat n_1 = d_1` and `vec t * hat n_2 = d_2`

Therefore, for all real values of λ, we have

` vec t * ( hat n_1 + lambda hat n_2) = d_1 + lambda d_2`

`=>` Since `vec t` is arbitrary, it satisfies for any point on the line.

`=>` Hence, the equation `vec r * ( hat n_1 + lambda hat n_2) = d_1 + lambda d_2` represents a plane `pi_31 ` which is such that if any vector `vec r` satisfies both the equations `π_1` and `π_2`, it also satisfies the equation `π_3` i.e., any plane passing through the intersection of the planes

`vec r * vec n_1 = d_1` and `vec r * vec n_2 = d_2`

has the equation `color{red}{vec r * (vec n_1 + lambda vec n_2) = d_1 +lambda d_2}` ............(1)

`color{brown} "Cartesian form"`

In Cartesian system, let

`vec n_1 = A_1 hat i + B_2 hat j + C_1 hat k`

`vec n_2 = A_2 hat i + B_2 hat j + C_2 hat k`

and `vec r = x hat i + y hat j + z hat k`

Then (1) becomes

`x (A_1 + λA_2) + y (B_1 + λB_2) + z (C_1 + λC_2) = d_1 + λd_2`

or`color{red}{ (A_1 x + B_1 y + C_1 z – d_1) + λ(A_2 x + B_2 y + C_2 z – d_2) = 0}` ... (2)

which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

`=>` If `vec t` is the position vector of a point on the line, then

`vec t * hat n_1 = d_1` and `vec t * hat n_2 = d_2`

Therefore, for all real values of λ, we have

` vec t * ( hat n_1 + lambda hat n_2) = d_1 + lambda d_2`

`=>` Since `vec t` is arbitrary, it satisfies for any point on the line.

`=>` Hence, the equation `vec r * ( hat n_1 + lambda hat n_2) = d_1 + lambda d_2` represents a plane `pi_31 ` which is such that if any vector `vec r` satisfies both the equations `π_1` and `π_2`, it also satisfies the equation `π_3` i.e., any plane passing through the intersection of the planes

`vec r * vec n_1 = d_1` and `vec r * vec n_2 = d_2`

has the equation `color{red}{vec r * (vec n_1 + lambda vec n_2) = d_1 +lambda d_2}` ............(1)

`color{brown} "Cartesian form"`

In Cartesian system, let

`vec n_1 = A_1 hat i + B_2 hat j + C_1 hat k`

`vec n_2 = A_2 hat i + B_2 hat j + C_2 hat k`

and `vec r = x hat i + y hat j + z hat k`

Then (1) becomes

`x (A_1 + λA_2) + y (B_1 + λB_2) + z (C_1 + λC_2) = d_1 + λd_2`

or`color{red}{ (A_1 x + B_1 y + C_1 z – d_1) + λ(A_2 x + B_2 y + C_2 z – d_2) = 0}` ... (2)

which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

Q 3157180084

Find the vector equation of the plane passing through the intersection of

the planes `vec r * (hat i + hat j + hat k) = 6` and ` vec r * (2 hat i + 3 hat i + 4 hat k) = -5` , and the point ` (1,1,1)`

Class 12 Chapter 11 Example 20

the planes `vec r * (hat i + hat j + hat k) = 6` and ` vec r * (2 hat i + 3 hat i + 4 hat k) = -5` , and the point ` (1,1,1)`

Class 12 Chapter 11 Example 20

Here, `vec (n_1) = hat i + hat j + hat k` and `vec ( n_2) = 2 hat i + 3 hat j + 4 hat k `;

and `d_1 = 6` and `d_2 = –5`

Hence, using the relation `vec r * (vec n_1 + lambda vec n_2) = d_1 = lambda d_2` , we get

`vec r * [ hat i + hat j + hat k + lambda (2 hat i + 3 hat j + 4 hat k) ] = 6 -5 lambda`

or `vec r * [ (1+ 2lambda) hat i + (1+ 3 lambda ) hat j + (1 + 4 lambda) hat k ] = 6 - 5 lambda` .......(1)

where, λ is some real number

Taking `vec r = x hat i + y hat j + z hat k` , we get

` (x hat i + y hat j + z hat k) * [ (1+ 2 lambda ) hat i + (1+ 3 lambda) hat j + (1+4 lambda) hat k ] = 6 - 5 lambda`

or (1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ

or (x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 ... (2)

Given that the plane passes through the point (1,1,1), it must satisfy (2), i.e.

(1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0

or `lambda = 3/14`

Putting the values of λ in (1), we get

`vec r [ (1+3/7) hat i + (1+9/14) hat j + (1+6/7) hat k] = 6 - 15/14`

or `vec r (10/7 hat i + 23/14 hat j + 13/7 hat k) = 69/14`

or `vec r * (20 hat i + 23 hat j + 26 hat k) = 69`

which is the required vector equation of the plane.

`=>` Let the given lines be

`vec r = vec a_1 + lambda vec b_1` .....(1)

and `vec r = vec a_2 + mu vec b_2` .....(2)

`=>` The line (1) passes through the point, say A, with position vector `vec a_1` and is parallel to`vec b_1`. The line (2) passes through the point, say B with position vector `vec a_2` and is parallel to `vec b_2` .

Thus , `bar (AB) = vec a_2 - vec a_1`

`=>` The given lines are coplanar if and only if `bar (AB)` is perpendicular to `vec b_1 xx vec b_2`.

i.e., `bar (AB) * (vec b_1 xx vec b_2) = 0` or `(vec a_2 - vec a_1)* ( vec b_1 xx vec b_2) =0`

`color{brown} "Cartesian form"`

`=>` Let `(x_1, y_1, z_1)` and `(x_2, y_2, z_2)` be the coordinates of the points A and B respectively.

`=>` Let `a_1, b_1, c_1` and `a_2, b_2, c_2` be the direction ratios of `vec b_1` and `vec b_2` , respectively. Then

`bar (AB) = (x_2 - x_1) hat i + (y_2 -y_1) hat j + (z_2 - z_1) hat k`

`vec b_1 = a_1 hat i + b_1 hat j + c_1 hat k` and `vec b_2 = a_2 hat i + b_2 hat j + c_2 hat k`

`=>` The given lines are coplanar if and only if `bar (AB) * (vec b_1 xx vec b_2) =0` . In the cartesian form, it can be expressed as

`color{red}{ | ( x_2 -x_1 , y_2- y_1 , z_2- z_1 ) , ( a_1 , b_2 ,c_1) , (a_2, b_2, c_2) | = 0}` ........ (4)

`vec r = vec a_1 + lambda vec b_1` .....(1)

and `vec r = vec a_2 + mu vec b_2` .....(2)

`=>` The line (1) passes through the point, say A, with position vector `vec a_1` and is parallel to`vec b_1`. The line (2) passes through the point, say B with position vector `vec a_2` and is parallel to `vec b_2` .

Thus , `bar (AB) = vec a_2 - vec a_1`

`=>` The given lines are coplanar if and only if `bar (AB)` is perpendicular to `vec b_1 xx vec b_2`.

i.e., `bar (AB) * (vec b_1 xx vec b_2) = 0` or `(vec a_2 - vec a_1)* ( vec b_1 xx vec b_2) =0`

`color{brown} "Cartesian form"`

`=>` Let `(x_1, y_1, z_1)` and `(x_2, y_2, z_2)` be the coordinates of the points A and B respectively.

`=>` Let `a_1, b_1, c_1` and `a_2, b_2, c_2` be the direction ratios of `vec b_1` and `vec b_2` , respectively. Then

`bar (AB) = (x_2 - x_1) hat i + (y_2 -y_1) hat j + (z_2 - z_1) hat k`

`vec b_1 = a_1 hat i + b_1 hat j + c_1 hat k` and `vec b_2 = a_2 hat i + b_2 hat j + c_2 hat k`

`=>` The given lines are coplanar if and only if `bar (AB) * (vec b_1 xx vec b_2) =0` . In the cartesian form, it can be expressed as

`color{red}{ | ( x_2 -x_1 , y_2- y_1 , z_2- z_1 ) , ( a_1 , b_2 ,c_1) , (a_2, b_2, c_2) | = 0}` ........ (4)

Q 3147580483

Show that the lines

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

Class 12 Chapter 11 Example 21

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

Class 12 Chapter 11 Example 21

Here, `x_1 = – 3, y_1 = 1, z_1 = 5, a_1 = – 3, b_1 = 1, c_1 = 5`

`x_2 = – 1, y_2 = 2, z_2 = 5, a_2 = –1, b_2 = 2, c_2 = 5`

Now, consider the determinant

`| (x_2 -x_1 , y_2-y_1 , z_2-z_1 ), (a_1 ,b_1 , c_1 ), (a_2, b_2 ,c_2) | = | (2,1,0), (-3,1,5), (-1,2,5) | = 0`

Therefore, lines are coplanar.

`=>` The angle between two planes is defined as the angle between their normals (Fig (a)).

`=>` If `θ` is an angle between the two planes, then so is `180 – θ` (Fig (b)). We shall take the acute angle as the angles between two planes.

`=>` If `vec n_1` and `vec n_2` are normals to the planes and θ be the angle between the planes

`vec r * vec n_1 = d_1` and `vec r * vec n_2 = d_2` .

`=>` Then `θ` is the angle between the normals to the planes drawn from some common point.

We have, `color{red}{cos theta = | (vec n_1 *vec n_2 )/( | vec n_1 | | vec n_2 |) |}`

`color{brown} "Cartesian form"`

`=>` Let `θ` be the angle between the planes,

`A_1 x + B_1 y + C_1z + D_1 = 0` and `A_2x + B_2 y + C_2 z + D_2 = 0`

The direction ratios of the normal to the planes are `A_1, B_1, C_1` and `A_2, B_2, C_2` respectively

Therefore,`color{red}{ cos θ = | ( 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) ) |}`

`=>` If `θ` is an angle between the two planes, then so is `180 – θ` (Fig (b)). We shall take the acute angle as the angles between two planes.

`=>` If `vec n_1` and `vec n_2` are normals to the planes and θ be the angle between the planes

`vec r * vec n_1 = d_1` and `vec r * vec n_2 = d_2` .

`=>` Then `θ` is the angle between the normals to the planes drawn from some common point.

We have, `color{red}{cos theta = | (vec n_1 *vec n_2 )/( | vec n_1 | | vec n_2 |) |}`

`color{red} "Key Point "` - The planes are perpendicular to each other if `vec n_1 * vec n_2 =0` and parallel if `vec n_1` is parallel to `vec n_2`

`color{brown} "Cartesian form"`

`=>` Let `θ` be the angle between the planes,

`A_1 x + B_1 y + C_1z + D_1 = 0` and `A_2x + B_2 y + C_2 z + D_2 = 0`

The direction ratios of the normal to the planes are `A_1, B_1, C_1` and `A_2, B_2, C_2` respectively

Therefore,`color{red}{ cos θ = | ( 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) ) |}`

`color{red} "Key Points "`

1. If the planes are at right angles, then `θ = 90^o` and so `cos θ = 0.`

Hence,` color{blue}{cos θ = A_1A_2 + B_1B_2 + C_1C_2 = 0}`.

2. If the planes are parallel, then `color{blue}{ (A_1)/(A_2) = (B_1)/(B_2) = (C_1)/(C_2)}`

1. If the planes are at right angles, then `θ = 90^o` and so `cos θ = 0.`

Hence,` color{blue}{cos θ = A_1A_2 + B_1B_2 + C_1C_2 = 0}`.

2. If the planes are parallel, then `color{blue}{ (A_1)/(A_2) = (B_1)/(B_2) = (C_1)/(C_2)}`

Q 3157580484

Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7

using vector method.

Class 12 Chapter 11 Example 22

using vector method.

Class 12 Chapter 11 Example 22

The angle between two planes is the angle between their normals. From the

equation of the planes, the normal vectors are

`vec (N_1) = 2 hat i + hat j -2 hat k` and `vec (N_2 ) = 3 hat i - 6 hat j -2 hat k`

Therefore ` cos theta = | (vec (N_1 ) * vec (N_2) )/( | vec N_1 | | vec N_2 |) | = | ( (2 hat i + hat j -2 hat k) * ( 3 hat i - 6 hat j - 2hat k) )/( sqrt (4+1+4) sqrt (9+36+4) ) | = (4/21)`

Hence `theta = cos^(-1) (4/21)`

Q 3117680580

Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.

Class 12 Chapter 11 Example 23

Class 12 Chapter 11 Example 23

Comparing the given equations of the planes with the equations

`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`

We get `A_1 = 3, B_1 = – 6, C_1 = 2`

`A_2 = 2, B_2 = 2, C_2 = – 2`

`cos theta = | (3 xx 2 + (-6) (2) + (2) (-2) )/( sqrt (3^2 + (-6)^2 + (-2)^2 ) sqrt ( 2^2 + 2^2 + (-2)^2) ) | `

`= | (-10)/(7 xx 2 sqrt 3 ) | = 5/(7 sqrt 3) = (5 sqrt 3)/21`

Therefore, ` theta = cos^(-1) ( (5 sqrt 3)/(21) )`

`"Vector form"`

`=>` Consider a point P with position vector `vec a` and a plane `π_1` whose equation is `vec r * hat n = d` (Fig).

`=>` Consider a plane `π_2` through P parallel to the plane `π_1`. The unit vector normal to `π_2` is `hat n` Hence, its equation is ` (vec r - vec a) * hat n = 0`

i.e., `vec r * hat n = vec a * hat n`

`=>` Thus, the distance ON′ of this plane from the origin is ` | vec a * hat n |` . Therefore, the distance PQ from the plane `π_1` is (Fig. (a))

i.e.,

`=>` which is the length of the perpendicular from a point to the given plane.

`\color{green} ✍️` We may establish the similar results for (Fig 11.19 (b)).

`color{brown} "Cartesian form"`-

`=>` Let `P(x_1, y_1, z_1) `be the given point with position vector `vec a` and

`Ax + By + Cz = D` be the Cartesian equation of the given plane. Then

`vec a = x_1 hat i + y_1 hat j + z_1 hat k`

`bar N = A hat i + B hat j + C hat k`

Hence, from Note 1, the perpendicular distance from P to the plane is

` color{blue}{d = | ( (x_1 hat i + y_1 hat j + z_1 hat k ) * ( A hat i + b hat j + C hat k) -D )/(sqrt (A^2 +B^2 +C^2 ) ) |}`

`=>` Consider a point P with position vector `vec a` and a plane `π_1` whose equation is `vec r * hat n = d` (Fig).

`=>` Consider a plane `π_2` through P parallel to the plane `π_1`. The unit vector normal to `π_2` is `hat n` Hence, its equation is ` (vec r - vec a) * hat n = 0`

i.e., `vec r * hat n = vec a * hat n`

`=>` Thus, the distance ON′ of this plane from the origin is ` | vec a * hat n |` . Therefore, the distance PQ from the plane `π_1` is (Fig. (a))

i.e.,

`ON - ON ' = | d - vec a * hat n |`

`=>` which is the length of the perpendicular from a point to the given plane.

`\color{green} ✍️` We may establish the similar results for (Fig 11.19 (b)).

`color{red} "Key Points "` -

1. If the equation of the plane `π_2` is in the form `vec r * bar N = d` , where `bar N` is normal to the plane, then the perpendicular distance is `color{blue}{ ( vec a * bar N -d )/( | bar N| )}.`

2. The length of the perpendicular from origin O to the plane `color{blue}{vec r * bar N= d}` is ` ( |d| )/(| bar N | )`

(since `vec a = 0` )

1. If the equation of the plane `π_2` is in the form `vec r * bar N = d` , where `bar N` is normal to the plane, then the perpendicular distance is `color{blue}{ ( vec a * bar N -d )/( | bar N| )}.`

2. The length of the perpendicular from origin O to the plane `color{blue}{vec r * bar N= d}` is ` ( |d| )/(| bar N | )`

(since `vec a = 0` )

`color{brown} "Cartesian form"`-

`=>` Let `P(x_1, y_1, z_1) `be the given point with position vector `vec a` and

`Ax + By + Cz = D` be the Cartesian equation of the given plane. Then

`vec a = x_1 hat i + y_1 hat j + z_1 hat k`

`bar N = A hat i + B hat j + C hat k`

Hence, from Note 1, the perpendicular distance from P to the plane is

` color{blue}{d = | ( (x_1 hat i + y_1 hat j + z_1 hat k ) * ( A hat i + b hat j + C hat k) -D )/(sqrt (A^2 +B^2 +C^2 ) ) |}`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red}{d = | (A x_1 + B y_1 +C z_1 -D )/( sqrt (A^2 +B^2 + C^2) ) |}`

Q 3147680583

Find the distance of a point (2, 5, – 3) from the plane

`vec r * ( 6 hat i − 3 hat j + 2 hat k ) = 4`

Class 12 Chapter 11 Example 24

`vec r * ( 6 hat i − 3 hat j + 2 hat k ) = 4`

Class 12 Chapter 11 Example 24

Here, `vec a = 2 hat i + 5 hat j -3 hat k , vec N = 6 hat i -3 hat j +2 hat k` and `d =4`

Therefore, the distance of the point (2, 5, – 3) from the given plane is

`( | ( 2 hat i + 5 hat j -3 hat k) * ( 6 hat i -3 hat j +2 hat k -4) -4 | )/( | 6 hat i -3 hat j +2 hat k | ) = ( | 12-15-6-4)/( sqrt (36 + 9 + 4) ) = 13/7`

`=>` The angle between a line and a plane is the complement of the angle between the line and normal to the plane (Fig).

`"Vector form"` If the equation of the line is `vec r = vec a + lambda vec b` and the equation of the plane is

`vec r * vec n = d`. Then the angle θ between the line and the normal to the plane is

`color{red}{cos theta = | (vec b * vec n )/( | vec b| * | vec n| )|}`

and so the angle `φ` between the line and the plane is given by `90 – θ,` i.e.,

`sin (90 – θ) = cos θ`

i.e.,

`"Vector form"` If the equation of the line is `vec r = vec a + lambda vec b` and the equation of the plane is

`vec r * vec n = d`. Then the angle θ between the line and the normal to the plane is

`color{red}{cos theta = | (vec b * vec n )/( | vec b| * | vec n| )|}`

and so the angle `φ` between the line and the plane is given by `90 – θ,` i.e.,

`sin (90 – θ) = cos θ`

i.e.,

` \ \ \ \ \ \ \ \ \ \ \ \sin φ = | (vec b * vec n)/( | vec b | |vec n| ) |` or `color{red}{ φ = sin^(-1) | (bar b * bar n )/( | bar b | | bar n ) |}`

Q 3127780681

Find the angle between the line

`(x+1)/2 = y/3 = (z-3)/6`

and the plane 10 x + 2y – 11 z = 3.

Class 12 Chapter 11 Example 25

`(x+1)/2 = y/3 = (z-3)/6`

and the plane 10 x + 2y – 11 z = 3.

Class 12 Chapter 11 Example 25

Let θ be the angle between the line and the normal to the plane. Converting the

given equations into vector form, we have

`vec r = (- hat i +3 hat k ) + lambda (2 hat i + 3 hat j+ 6 hat k )`

and `vec r * (10 hat i + 2 hat j -11 hat k ) = 3`

Here `vec b = 2 hat i +3 hat j + 6 hat k` and `vec n = 10 hat i +2 hat j -11 hat k`

`sin φ = | ( (2 hat i +3 hat j + 6 hat k )* (10 hat i +2 hat j -11 hat k ) )/( sqrt (2^2 +3^2 +6^2) sqrt ( 10^2 +2^2 + 11^2) ) |`

`= | (-40)/(7 xx 15) | = | (-8)/(21) | = 8/21` or `φ = sin^(-1) (8/21)`