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

• A. $\frac{π}{6}$
• B. $\frac{π}{4}$
• C. $\frac{π}{3}$
• D. $\frac{5π}{12}$

Answer 1

Answer: (C)

$\frac{π}{3}$

Answer 2

Let $P(x, y, z)$ be any point on plane $P_1$

Thereafter, $(x+4)^2+(y−2)^2+(z−1)^2=(x−2)^2+(y+2)^2+(z−3)^2$

$⇒12x−8y+4z+4=0$

$⇒3x−2y+z+1=0$

Also, $P_2 : 2x + y \+ 3z = 1$

$p_ 1\;and \; p_2$ create the angle of

$Cos θ = | \frac{6-2+3}{14}|$

$⇒ θ = \frac{π }{ 3}$

Hence, the correct option is (C): $\frac{π}{3}$