Question

The foot of perpendicular of the point $M(1,-2,1)$ on the plane $P:2x-y+2z+3=0$ is

• A. $(-3,-1,-3)$
• B. $(-2,2,-2)$
• C. $(-1,-1,-1)$
• D. $(-1,2,-3)$

Answer 1

The correct option is C $(-1,-1,-1)$

Given plane ,$P:2x-y+2z+3=0=ax+by+cz+d$
$D.r^{\prime }s$ of normal to the plane : $(a,b,c)=(2,-1,2)$
and point $M(1,-2,1)=(x_1,y_1,z_1)$
Let foot of perpendicular on the plane is $N(x_0,y_0,z_0)$
We know,
$\frac{x_0-x_1}{a}=\frac{y_0-y_1}{b}=\frac{z_0-z_1}{c}=-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}\Rightarrow \frac{x_0-1}{2}=\frac{y_0+2}{-1}=\frac{z_0-1}{2}=-\frac{2+2+2+3}{9}\Rightarrow x_0=-1,y_0=-1,z_0=-1$
So, point $N\equiv (-1,-1,-1).$