Question

In each of the following exercise, determine the direction cosines of the normal to plane and the distance from the origin:

z=2

x + y + z = 1

2x +3y -z = 5

5y + 8 = 0

Answer 1

Given, plane is z=2 or 0x + 0y +z=2.

The direction ratios of normal are 0,0 and 1.

Also, $\sqrt{0^2+0^2+1^2}=1$

Dividing both sides of Eq. (i) by 1, we obtain 0.x + 0.y + 1.z=2

Which is of the form lx+my+nz=d, where l,m,n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Hence, direction cosines are 0, 0 and 1 and the distance of the plane to the origin is 2 units.

Given, plane is x +y +z=1

The direction ratios of normal are 1,1 and 1.

Also, $\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Dividing both sides of Eq. (i) by $\sqrt{3}$, we obtain

$\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}$

Which is of the form lx +my +nz =d, where l,m,n are direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}and\frac{1}{\sqrt{3}}$ and the distance of normal from the origin is $\frac{1}{\sqrt{3}}$ units.

Given, equation of plane is 2x +3y -z = 5

The direction ratios of normal are 2,3 and -1.

Also $\sqrt{2^2+3^2+(-1{)}^2}=\sqrt{14}$

Dividing both sides of Eq. (i) by $\sqrt{14}$, we obtain

$\left(\frac{2}{\sqrt{14}}\right)x+\left(\frac{3}{\sqrt{14}}\right)y+(-\frac{1}{\sqrt{14}})z=\frac{5}{\sqrt{14}}$

Which is of the form lx +my +nz= d, where l,m,n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are

$\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}and-\frac{1}{\sqrt{14}}$ and the distance of normal to the origin is $\frac{5}{\sqrt{14}}$ units.

Given equation is 5y + 8 = 0 , which can be written as

$0x+(-1)y+0z=\frac{8}{5}$

Which is of the form lx +my +nz= d, where l,m,n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, -1 and 0 and distance of normal from the origin is 8/5 units.