Question

If foot of the perpendicular of P(2,-3,1) on the line
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+2}{-1}$is
$Q(a,b,c)$ then find the value of $-14(a+b+c)$

Answer 1

Given line is $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+2}{-1}=r...\left(1\right)$
Point $P\equiv (2,-3,1)$
Co-ordinates of foot of the perpendicular on line $\left(1\right)$ may be taken as $Q\equiv (2r-1,3r+3,-r-2)$
We get direction ratios of $PQ=2r-3,3r+6,-r-3$
Direction ratios of line segment are $2,3,-1$ from $\left(1\right)$
Since $PQ$ perpendicular to $AB$
$\therefore 2(2r-3)+3(3r+6)-1(-r-3)=0$
or,$14r+15=0$
$\therefore r=\frac{-15}{14}$
$\therefore Q\equiv (2r-1,3r+3,-r-2)=(\frac{-22}{7},\frac{-3}{14},\frac{-13}{14})$
$=(a,b,c)$ given
$\Rightarrow -14(a+b+c)=-14(\frac{-22}{7}+\frac{-13}{14}+\frac{-13}{14})=44+3+13=60$