JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q15

← JEE Main 2022 · 26 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 15 of 90

Let the plane $2x+3y+z+20=0$ be rotated through a right angle about its line of intersection with the plane $x-3y+5z=8$ . If the mirror image of the point $\left(2,-\dfrac12,2\right)$ in the rotated plane is B $(a,b,c)$ , then :

\dfrac a8=\dfrac b5=\dfrac c{-4}

\dfrac a4=\dfrac b5=\dfrac c{-2}

\dfrac a8=\dfrac b{-5}=\dfrac c4

\dfrac a4=\dfrac b5=\dfrac c2

Answer and worked solution

Official answer: A

Rotated plane $P_1+\lambda P_2=0$ perpendicular to $P_1$ : $2(2+\lambda)+3(3-3\lambda)+(1+5\lambda)=14-2\lambda=0$ , so $\lambda=7$ : $9x-18y+36z-36=0$ , i.e. $x-2y+4z=4$ .

Image of $\left(2,-\frac12,2\right)$ . Value $2+1+8-4=7$ , $|\vec n|^2=21$ , so the image is $\left(2,-\frac12,2\right)-\frac{14}{21}(1,-2,4)=\left(\frac43,\frac56,-\frac23\right)$ .

Ratios. $a:b:c=8:5:-4$ .

(A)

Topics:Equation of a plane Angles and distances

← Previous question Next question →

JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 15