Question

Find the equation of the plane passing through the point (1, 1, 1) and is perpendicular to the line `("x" - 1)/3 = ("y" - 2)/0 = ("z" - 3)/4`. Also, find the distance of this plane from the origin.

Answer 1

Since the plane is perpendicular to the given line, its direction ratios are proportional to 3, 0, 4.

So, the required equation of the plane is

3x + 0y + 4z + d = 0 ......(i)

where d is a constant.

Since this plane passes through (1, 1, 1)

3 + 0 + 4 + d = 0

⇒ d = − 7

From equation (i), we get

3x + 4z − 7 = 0 .........(ii)

This is the required equation of the plane.

Perpendicular distance of (ii) from the origin is given by,

d = `|"ax"_1 + "by"_1 + "cz"_1 + "d"|/sqrt("a"^2 + "b"^2 + "c"^2)`

⇒ d = `|3 xx 0 + 4 xx 0 - 7|/sqrt(3^2 + 0^2 + 4^2)`

⇒ d = `|-7|/sqrt(9 + 16)`

⇒ d = `7/sqrt25`

= `7/5` units