If from a point P (a, b, c) perpendiculars PA and PB are drawn to yz and zx-planes, then find the vector equation of the plane OAB.

Solution:

Point P is (a, b, c)
PA ⊥ y z-plane and PB ⊥ z x-plane.
∴ A is (0, b, c) and B is (a, 0, c)
We are to find the equation of plane through (0, 0, 0), (0, b, c) and (a, 0, c).
The equation of plane through (0, 0, 0) is
λ (x – 0) + μ (y – 0) + v (z – 0) = 0
∴ λx + μ y + v z = 0    ...(1)

∴ it passes through (0, b, c) and (a, 0, c)
∴ 0 λ + b μ + c v = 0
and a λ + 0 μ + c v = 0
Solving these, we get,

$\frac{λ}{bc-0}$ =   $\frac{µ}{ca-0}$ = $\frac{v}{ca-0}$

Thus, λ = kbc , µ = kca, v=-kab

Putting values of λ, μ,v in (1), we get,
k b c x + k c a y – k a b z = 0

Or, $\frac{x}{a}$ + $\frac{y}{b}$ - $\frac{z}{c}$.