JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q15

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MathematicsSingle correct+4 / -1Question 15 of 90

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

\dfrac\pi6

\dfrac\pi4

\dfrac\pi3

\dfrac{5\pi}{12}

Answer and worked solution

Official answer: C

Plane P is the perpendicular bisector plane, with normal along $(2,-2,3)-(-4,2,1)=(6,-4,2)\parallel(3,-2,1)$ .

Angle. $\cos\theta=\dfrac{(3,-2,1)\cdot(2,1,3)}{\sqrt{14}\sqrt{14}}=\dfrac7{14}=\dfrac12$ , so $\theta=\dfrac\pi3$ .

(C)

Topics:Equation of a plane Angles and distances

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 15