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)  x+y+z=1 
(c)2x+3y-z=5  (d)  5y+8=0

Answer 1

(a) The equation of the plane is z=2 or 0x+0y+z=2...(1)

The direction ratios of the normal are 0, 0, and 1.
∴\(\sqrt {0^2+0^2+1^2}=1\)

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

This 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.
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

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(b) x+y+z=1...(1)

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

∴\(\sqrt {(1)^2+(1)^2+(1)^2}=\sqrt 3\)

Dividing both sides of equation(1) by \(\sqrt 3\), we obtain
\(\frac{1}{\sqrt 3}x+\frac{1}{\sqrt 3}y+\frac{1}{\sqrt 3}z=\frac{1}{\sqrt 3}\)                                     ...(2)

This equation is 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 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.

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(c) 2x+3y-z=5...(1)

The direction ratios of normal are 2, 3, and -1.
∴\(\sqrt{(2)^2+(3)^2+(-1)^2}=\sqrt 14\)

Dividing both sides of equation(1) by 14, we obtain
\(\frac{2}{\sqrt {14}}x+\frac{3}{\sqrt{14}}y-\frac{1}{\sqrt {14}}z=\frac{5}{\sqrt {14}}\)

This equation 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 from the origin is \(\frac{5}{\sqrt{14}}\) units.

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(d) 5y+8=0
\(\Rightarrow \) 0x-5y+0z=8...(1)

The direction ratios of normal are 0, -5, and 0.
∴\(\sqrt{0+(-5)^2+0}\) =5

Dividing both sides of equation(1) by 5, we obtain
-y=\(\frac{8}{5}\)

This equation 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 the distance of normal from the origin is \(\frac{8}{5}\) units.