Question

The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is

• A. $\sqrt{17}$
• B. $\sqrt{29}$
• C. $\sqrt{34}$
• D. $\sqrt{41}$

Answer 1

The correct option is C $\sqrt{34}$

$P(1,2,3),$ Plane :$x-y+z=5$

$F$ is foot of perpendicular form $P$ to plane and $I$ is image,then $PF=FI$

$\therefore$ If $(x,y,z)=(r+1,-r+2,r+3)$ are foot of perpendicular.

$\Rightarrow (r+1)-(-r+2)+r+3=5\Rightarrow r=1\therefore F=(2,1,4)\therefore I=(3,0,5)$

$\therefore$ distance of $I$ from origin $=\sqrt{3^2+0^2+5^2}=\sqrt{34}$