jee-main 2024 Q79

jee-main · India · session1_27jan_shift1 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)

The distance, of the point $( 7 , - 2,11 )$ from the line $\frac { x - 6 } { 1 } = \frac { y - 4 } { 0 } = \frac { z - 8 } { 3 }$ along the line $\frac { x - 5 } { 2 } = \frac { y - 1 } { - 3 } = \frac { z - 5 } { 6 }$, is:
(1) 12
(2) 14
(3) 18
(4) 21

The distance, of the point $( 7 , - 2,11 )$ from the line $\frac { x - 6 } { 1 } = \frac { y - 4 } { 0 } = \frac { z - 8 } { 3 }$ along the line $\frac { x - 5 } { 2 } = \frac { y - 1 } { - 3 } = \frac { z - 5 } { 6 }$, is:\\
(1) 12\\
(2) 14\\
(3) 18\\
(4) 21

Paper Questions

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