jee-main 2024 Q90

jee-main · India · session2_05apr_shift2 Vectors: Lines & Planes Distance Computation (Point-to-Plane or Line-to-Line)

Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$

Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$

Paper Questions

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