question_answer

The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is : [JEE Main Solved Paper-2016 ]

A) \[\frac{20}{3}\]                                

B) \[3\sqrt{10}\]

C) \[10\sqrt{3}\]                   

D) \[\frac{10}{\sqrt{3}}\]

Correct Answer: C

Solution :

Equation of line parallel to x = y = z through (1, ?5, 9) is\[\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda \] If \[P(\lambda +1,\lambda -5\lambda +9)\] be point of intesection of line and plane. \[\Rightarrow \]\[\lambda +1-\lambda +5+\lambda +9=5\] \[\Rightarrow \]\[\lambda =-10\] \[\Rightarrow \]Coordinates point are (?9, ?15, ?1) \[\Rightarrow \]Required distance \[=10\sqrt{3}\]