Question

The foot of the perpendicular from $P(1,0,2)$ to the line $\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1}$ is the point

• A. $(1,2,-3)$
• B. $(\frac{1}{2},1,-\frac{3}{2})$
• C. $(2,4,-6)$
• D. $(2,3,6)$

Answer 1

Answer: (B)

$(\frac{1}{2},1,-\frac{3}{2})$

$\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1}=t$

Let the foot of the perpendicular on the line Is $T(3t-1,-2t+2,-r-1)$

and point $P^{\prime }(-1,2,-1)$ is present on the line.

DR's of $P^{\prime }T=3,-2,-1$

and DR's of $PT=3t-2,-2t+2,-t-3$

$P^{\prime }T\perp PT\Rightarrow 9t-6+4t-4+t+3=0\Rightarrow t=1/2$

Foot of the perpendicular is $:(\frac{1}{2},1,-\frac{3}{2})$