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

Updated On: Sep 19, 2023
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Solution and Explanation

(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.
02+02+12=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.


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

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

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

Dividing both sides of equation(1) by 3\sqrt 3, we obtain
13x+13y+13z=13\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 13,13\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}, and 13\frac{1}{\sqrt 3} and the distance of normal from the origin is 13\frac{1}{\sqrt 3} units.


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

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

Dividing both sides of equation(1) by 14, we obtain
214x+314y114z=514\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 214,314\frac{2}{\sqrt {14}},\frac{3}{\sqrt {14}}, and 114\frac{-1}{\sqrt{14}} and the distance of normal from the origin is 514\frac{5}{\sqrt{14}} units.


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

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

Dividing both sides of equation(1) by 5, we obtain
-y=85\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 85\frac{8}{5} units.

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Concepts Used:

Plane

A  surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely:

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Properties of a Plane:

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