JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q16

MathematicsSingle correct+4 / -1Question 16 of 90

The foot of the perpendicular from a point on the circle $x^2+y^2=1$ , $z=0$ to the plane $2x+3y+z=6$ lies on which one of the following curves ?

(6x+5y-12)^2+4(3x+7y-8)^2=1

z=6-2x-3y

(5x+6y-12)^2+4(3x+5y-9)^2=1

z=6-2x-3y

(6x+5y-14)^2+9(3x+5y-7)^2=1

z=6-2x-3y

(5x+6y-14)^2+9(3x+7y-8)^2=1

z=6-2x-3y

Answer and worked solution

Official answer: B

Foot of $(\cos t,\sin t,0)$ on $2x+3y+z=6$ : subtract $\lambda(2,3,1)$ with $\lambda=\frac{2\cos t+3\sin t-6}{14}$ .

Eliminate t. Then $5x+6y-12=\cos t$ and $3x+5y-9=\frac12\sin t$ , so $(5x+6y-12)^2+4(3x+5y-9)^2=1$ on the plane.

(B)

Topics:Equation of a plane Angles and distances

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 16