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

Answer: (B)

The dr’s of PA are $x_1-1,y_1-3,z_1-4$
The dr’s $\overrightarrow{n}$ are 2, -1, 1
The dr’s of PA and $\overrightarrow{n}$ are parallel
$\therefore \frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\lambda$
$x_1=2\lambda +1,y_1=-\lambda +3,z_1=\lambda +4$
$\therefore A=\left(2\lambda +1,-\lambda +3,\lambda +4\right)lieson2x-y+z+3=0$
$\Rightarrow -4\lambda +2-\lambda -3-\lambda +4+3=0$
$6\lambda =-6$
$\lambda =-1$
$\therefore P=\left(-1,4,3\right)$