The co-ordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line…
The co-ordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line
$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ are
$(-2,-3,1)$
$(2,3,-1)$
$(2,3,1)$
(-2,-3,-1)
Solution
Let $\alpha$ be the foot of the Ler drawn from the point $\mathrm{P}(0,2,3)$ to the given line.
Co-ordinates of any point on given line are $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$...say
Let $Q \equiv(5 \lambda-3,2 \lambda+1,3 \lambda-4)$ and $P \equiv(0,2,3)$
$\therefore$ d.r. of PQ are $5 \lambda-3,2 \lambda-1,3 \lambda-7$
d.r. of given line are $5,2,3$
$\therefore 5(5 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-7)=0$
$\Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \Rightarrow 38 \lambda-38=0 \Rightarrow \lambda=1$
$\therefore \mathrm{Q} \equiv(2,3,-1)$
