Question

The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane

• A. $x-y-2z=1$
• B. $x-2y+z=1$
• C. $2x+y-z=1$
• D. $x+2y-z=1$

Answer 1

The correct option is C $2x+y-z=1$

Let $P=(4,2,3),A=(1,-2,3)$ and $B=(1,1,0)$


Equation of line $AB$ is
$\frac{x-1}{1-1}=\frac{y+2}{1+2}=\frac{z-3}{0-3}=\lambda \text{(say)}$
Point $M$ is $(1,3\lambda -2,-3\lambda +3)$

$\text{D.r's of}PM=(1-4,3\lambda -2-2,-3\lambda +3-3)$
$\Rightarrow \text{D.r's of}PM=(-3,3\lambda -4,-3\lambda )$ and
$\text{D.r's of}AB=(0,3,-3)$

Since $PM\perp AB$
$(-3)\cdot 0+(3\lambda -4)\cdot \left(3\right)+(-3\lambda )\cdot (-3)=0$
$\Rightarrow 9\lambda -12+9\lambda =0$
$\Rightarrow 18\lambda -12=0\Rightarrow \lambda =\frac{2}{3}$
$\therefore M=(1,0,1)$
Clearly, $M$ lies on plane $2x+y-z=1$