Question

Find the equation of the plane passing through the points (2, 3, 1), (4, –5, 3) and parallel to x-axis.

Answer 1

We know that the equation of a plane parallel to x-axis is

by + cz + d = 0  (∵ a = 0)   ...(i)

Given, it passes through the points (2, 3, 1) and (4, –5, 3).

We get 3b + c + d = 0   ...(ii)

And –5b + 3c + d = 0   ...(iii)

Solving equations (ii) and (iii)

`b/(1 - 3) = c/(-5 - 3) = d/(9 - (-5))`

⇒ `b/(-2) = c/(-8) = d/14`

⇒ `b/(-1) = c/(-4) = d/7 = λ`   ...(Say)

∴ b = –λ, c = –4λ, d = 7λ

Substitute these values in equation (i), we get

–λy – 4λz + 7λ = 0

y + 4z – 7 = 0

∴ Equation of plane is y + 4z = 7.