jee-main 2022 Q78

jee-main · India · session1_27jun_shift2 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)

Let the foot of the perpendicular from the point $( 1,2,4 )$ on the line $\frac { x + 2 } { 4 } = \frac { y - 1 } { 2 } = \frac { z + 1 } { 3 }$ be $P$. Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$ is
(1) $\frac { 50 } { 13 }$
(2) $\frac { 63 } { 13 }$
(3) $\frac { 65 } { 13 }$
(4) 4

Let the foot of the perpendicular from the point $( 1,2,4 )$ on the line $\frac { x + 2 } { 4 } = \frac { y - 1 } { 2 } = \frac { z + 1 } { 3 }$ be $P$. Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$ is\\
(1) $\frac { 50 } { 13 }$\\
(2) $\frac { 63 } { 13 }$\\
(3) $\frac { 65 } { 13 }$\\
(4) 4

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