jee-main 2019 Q88

jee-main · India · session2_09apr_shift2 Vectors 3D & Lines Distance from a Point to a Line (Show/Compute)

The vertices $B$ and $C$ of a $\triangle A B C$ lie on the line, $\frac { x + 2 } { 3 } = \frac { y - 1 } { 0 } = \frac { z } { 4 }$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given the point $A ( 1 , - 1,2 )$, is
(1) 6
(2) $2 \sqrt { 34 }$
(3) $\sqrt { 34 }$
(4) $5 \sqrt { 17 }$

The vertices $B$ and $C$ of a $\triangle A B C$ lie on the line, $\frac { x + 2 } { 3 } = \frac { y - 1 } { 0 } = \frac { z } { 4 }$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given the point $A ( 1 , - 1,2 )$, is\\
(1) 6\\
(2) $2 \sqrt { 34 }$\\
(3) $\sqrt { 34 }$\\
(4) $5 \sqrt { 17 }$

Paper Questions

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