jee-main 2024 Q80

jee-main · India · session2_04apr_shift2 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)

Let P be the point of intersection of the lines $\frac { x - 2 } { 1 } = \frac { y - 4 } { 5 } = \frac { z - 2 } { 1 }$ and $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 2 }$. Then, the shortest distance of P from the line $4 x = 2 y = z$ is
(1) $\frac { 5 \sqrt { 14 } } { 7 }$
(2) $\frac { 3 \sqrt { 14 } } { 7 }$
(3) $\frac { \sqrt { 14 } } { 7 }$
(4) $\frac { 6 \sqrt { 14 } } { 7 }$

Let P be the point of intersection of the lines $\frac { x - 2 } { 1 } = \frac { y - 4 } { 5 } = \frac { z - 2 } { 1 }$ and $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 2 }$. Then, the shortest distance of P from the line $4 x = 2 y = z$ is\\
(1) $\frac { 5 \sqrt { 14 } } { 7 }$\\
(2) $\frac { 3 \sqrt { 14 } } { 7 }$\\
(3) $\frac { \sqrt { 14 } } { 7 }$\\
(4) $\frac { 6 \sqrt { 14 } } { 7 }$

Paper Questions

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