Question

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a)z = 2 (b) (c) (d)5 y + 8 = 0

Answer 1

The Cartesian equation of plane in normal form is expressed as,

lx+my+nz=d (1)

The direction cosines of the normal to the plane are,

l l 2 + m 2 + n 2  along x direction (2)

m l 2 + m 2 + n 2  along the y direction(3)

n l 2 + m 2 + n 2  along the z direction(4)

And, the distance from the origin is,

| d l 2 + m 2 + n 2 |(5)

(a)

The equation of plane is,

z=0 \(6)

Comparing equation (6) with equation (1), we get,

l=0

m=0

n=1

d=2

Substitute the values in equation (1), and then the direction cosine of the normal is

0 0 2 + 0 2 +1 =0 along the x direction

0 0 2 + 0 2 + 1 2 =0 along the y direction

1 0 2 + 0 2 + 1 2 =1 along the z direction

The distance from the origin is obtained by substituting the values of l,m,n and d in equation (5).

2 0 2 + 0 2 + 1 2 =2 units

Hence, the direction cosines of the normal to the plane are 0 in the x direction, 0 in the y directionand 1 in the z directionand the distance from the origin is 2 units.

(b)

The equation of plane is,

x+y+z=1 (7)

Comparing equation (7) with equation (1), we get,

l=1

m=1

n=1

d=1

Substitute the values in equation (1), and then calculate the direction cosine of the normal.

1 1 2 + 1 2 + 1 2 = 1 3  along the x direction

1 1 2 + 1 2 + 1 2 = 1 3  along the y direction

1 1 2 + 1 2 + 1 2 = 1 3  along the z direction

The distance from the origin is obtained by substituting the values of l,m,n and d in equation (5).

1 1 2 + 1 2 + 1 2 = 1 3  units

Hence, the direction cosines of the normal to the plane are 0 in the x direction, 0 in the y directionand 1 in the z directionand the distance from the origin is 1 3  units.

(c)

The equation of plane is,

2x+3y−z=5 (8)

Comparing equation (8) with equation (1), we get,

l=2

m=3

n=−1

d=5

Substitute the values in equation (1), and then calculate the direction cosine of the normal.

2 2 2 + 3 2 + ( −1 ) 2 = 2 4+9+1 = 2 14  along the x direction

3 2 2 + 3 2 + ( −1 ) 2 = 3 4+9+1 = 3 14  along the y direction

−1 2 2 + 3 2 + ( −1 ) 2 = −1 4+9+1 = −1 14  along the z direction

The distance from the origin is obtained by substituting the values of l,m,n and d in equation (5).

5 2 2 + 3 2 + ( −1 ) 2 = 5 4+9+1 = 5 14  units

Hence, the direction cosines of the normal to the plane are 0 in the x direction, 0 in the y direction and 1 in the z directionand the distance from the origin is 1 3  units.

(c)

The equation of plane is,

5y+8=0 5y=−8 (9)

Comparing equation (9) with equation (1), we get,

l=0

m=5

n=0

d=−8

Substitute the values in equation (1), and then calculate the direction cosine of the normal.

0 0 2 + 5 2 + 0 2 =0 along the x direction

5 0 2 + 5 2 + 0 2 =1 along the y direction

0 0 2 + 5 2 + 0 2 =0 along the z direction

The distance from the origin is obtained by substituting the values of l,m,n and d in equation (5).

| −8 0 2 + ( 5 ) 2 + 0 2 |= 8 5  units

Hence, the direction cosines of the normal to the plane are 0 in the x direction, 0 in the y directionand 1 in the z directionand the distance from the origin is 8 5  units.