Question

Foot of the perpendicular drawn from the point $(1,3,4)$ to the plane $2x–y+z+3=0$ is

• A. $(1,2,-3)$
• B. $(-1,4,3)$
• C. $(-3,5,2)$
• D. $(0,-4,-7)$

Answer 1

The correct option is B $(-1,4,3)$

Let $A$ be the given point and P$(x_1,y_1,z_1$) be the foot of perpendicular drawn
Then DR's of PA are $x_1–1,y_1–3,z_1–4$
The DR's of plane $(\overrightarrow{n}$) are $2,-1,1$
The DR’s of PA and $\overrightarrow{n}$ are parallel
$\therefore \frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\lambda x_1=2\lambda +1,y_1=-\lambda +3,z_1=\lambda +4\therefore P=(2\lambda +1,-\lambda +3,\lambda +4)$ lies on
$2x–y+z+3=0$
$\Rightarrow 4\lambda +2+\lambda –3+\lambda +4+3=0$ $6\lambda =-6$
$\lambda =-1$
$\therefore P=(-1,4,3)$