jee-main 2015 Q88

jee-main · India · 04apr Vectors: Lines & Planes Find Intersection of a Line and a Plane

The distance of the point $( 1,0,2 )$ from the point of intersection of the line $\frac { x - 2 } { 3 } = \frac { y + 1 } { 4 } = \frac { z - 2 } { 12 }$ and the plane $x - y + z = 16$, is
(1) 13
(2) $2 \sqrt { 14 }$
(3) 8
(4) $3 \sqrt { 21 }$

The distance of the point $( 1,0,2 )$ from the point of intersection of the line $\frac { x - 2 } { 3 } = \frac { y + 1 } { 4 } = \frac { z - 2 } { 12 }$ and the plane $x - y + z = 16$, is\\
(1) 13\\
(2) $2 \sqrt { 14 }$\\
(3) 8\\
(4) $3 \sqrt { 21 }$

Paper Questions

Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q70 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q86 Q87 Q88 Q89 Q90