Question

Find the equation of plane which is at a distance of 8 units from the origin and which is normal to the line having direction ratios 2,1,2 ?

• A. x + 2y + z = 24
• B. 2x + y + 2z = 24
• C. 2x + 2y + z = 72
• D. x + 2y + 2z = 72

Answer 1

The correct option is B

2x + y + 2z = 24

Here we are given the distance of plane from origin and the direction ratios of normal to the plane - So we can use equation of plane in normal form which is ,

lx + my + nz = d

(where l,m,n are the direction cosines of the normal to the plane and “d “ is the distance of the plane from origin.)

In the question we are given the direction ratios of normal and not the direction cosines -

So we’ll calculate direction cosines of normal first which will be - $\frac{2}{\sqrt{2^2+1^1+2^2}},\frac{1}{\sqrt{2^2+1^1+2^2}},\frac{2}{\sqrt{2^2+1^1+2^2}}$

Or $\frac{2}{3},\frac{1}{3},\frac{2}{3}$

So now we know l , m and n and “d” is already given in the question.

The equation of line will be - $\frac{2}{3}x+\frac{1}{3}y+\frac{2}{3}z=8$

Or 2x + y + 2z = 24