JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q13

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MathematicsSingle correct+4 / -1Question 13 of 90

Let the foot of the perpendicular from the point $(1,2,4)$ on the line $\dfrac{x+2}4=\dfrac{y-1}2=\dfrac{z+1}3$ be $P$ . Then the distance of $P$ from the plane $3x+4y+12z+23=0$ is

\dfrac{50}{13}

\dfrac{63}{13}

Answer and worked solution

Official answer: A

Foot. Point $(-2+4t,1+2t,-1+3t)$ ; perpendicularity gives $29-29t=0$ , $t=1$ : $P(2,3,2)$ .

Distance. $\dfrac{6+12+24+23}{13}=5$ .

(A)

Topics:Angles and distances Equation of a line in space

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 13