question_answer

If the perpendicular distance of the point (2, 3, 1) from the line \[3x+2y+z=0=x+2y\]is \[\frac{N}{\sqrt{2}}\] . Then N equals

A) 1                 

B)                   2

C) 4                             

D)    5

Correct Answer: D

Solution :

                Line is \[\frac{x}{1}=\frac{y}{-2}=\frac{z}{1}\] \[d=\frac{|\vec{v}\times \vec{c}|}{|\vec{c}|}=\frac{\sqrt{|\vec{v}{{|}^{2}}|\vec{c}{{|}^{2}}-(\vec{v}}.\vec{c}{{)}^{2}}}{|\vec{c}|}\] \[d=\frac{\sqrt{14\times 6-9}}{\sqrt{6}}=\frac{5\sqrt{3}}{\sqrt{6}}=\frac{5}{\sqrt{2}}\] \[\therefore \,\,N=5\]