Question

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

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

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

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