jee-main 2025 Q1

jee-main · India · session1_23jan_shift2 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)

The distance of the line $\frac { x - 2 } { 2 } = \frac { y - 6 } { 3 } = \frac { z - 3 } { 4 }$ from the point $( 1,4,0 )$ along the line $\frac { x } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { 3 }$ is :
(1) $\sqrt { 17 }$
(2) $\sqrt { 15 }$
(3) $\sqrt { 14 }$
(4) $\sqrt { 13 }$

The distance of the line $\frac { x - 2 } { 2 } = \frac { y - 6 } { 3 } = \frac { z - 3 } { 4 }$ from the point $( 1,4,0 )$ along the line $\frac { x } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { 3 }$ is :\\
(1) $\sqrt { 17 }$\\
(2) $\sqrt { 15 }$\\
(3) $\sqrt { 14 }$\\
(4) $\sqrt { 13 }$

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