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

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

Answer 1

The correct option is B $(\frac{1}{2},1,-\frac{3}{2})$

$\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1}=r$
$\therefore M(3r-1,-2r+2,-r-1)$
DR's of PM are $3r-2,-2r+2,-r-3$
DR's of line QM are $3,-2,-1$
$\therefore 9r-6+4r-4+r+3=0$
$\Rightarrow 14r-7=0\Rightarrow r=\frac{1}{2}$
$\therefore M(\frac{1}{2},1,-\frac{3}{2})$