Mathematics Different forms of equation of plane

### Topic Covered

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

### 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.

### Equation of a plane in normal form

=> 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.

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
Solution:

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
Solution:

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
Solution:

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
Solution:

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).

### Equation of a plane perpendicular to a given vector and passing through a given point

=> 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 vectorvec a and perpendicular to the vectorbar 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
Solution:

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.

### Equation of a plane passing through three non collinear points

=> 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.

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
Solution:

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

### Intercept form of the equation of a plane

=> 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.
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
Solution:

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

### Plane passing through the intersection of two given planes

=> 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

orcolor{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
Solution:

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.

### Coplanarity of Two Lines

=> 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 tovec 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
Solution:

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.

### Angle between Two Planes

=> 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{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)}

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
Solution:

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
Solution:

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) )

### Distance of a Point from a Plane

"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.,
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 )

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
Solution:

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

### Angle between a Line and a Plane

=> 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.,
 \ \ \ \ \ \ \ \ \ \ \ \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
Solution:

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)