Let the points on the plane P be equidistant from the points (-4,2,1) and (2,-2,3).

Question

Let the points on the plane $P$ be equidistant from the points $(-4,2,1)$ and $(2,-2,3)$. Then the acute angle between the plane $P$ and the plane $2x+y+3z=1$ is:

Accepted Answer

Let P(x, y, z) be any point on plane P1

Then $${(x + 4)^2} + {(y - 2)^2} + {(z - 1)^2} = {(x - 2)^2} + {(y + 2)^2} + {(z - 3)^2}$$

$$ \Rightarrow 12x - 8y + 4z + 4 = 0$$

$$ \Rightarrow 3x - 2y + z + 1 = 0$$

And ${P_2}:2x + y + 3z = 0$

$$\therefore$$ angle between P1 and P2

$$\cos \theta =\left| {{{6 - 2 + 3} \over {14}}} \right| \Rightarrow \theta  = {\pi  \over 3}$$

Solutions · 1