The foot of the perpendicular from the point $(1,2,3)$ on the line $\overline{\mathrm{r}}=(6…
The foot of the perpendicular from the point $(1,2,3)$ on the line $\overline{\mathrm{r}}=(6 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ has the co-ordinates
$(3,5,9)$
$(5,-3,9)$
$(3,-5,-9)$
$(5,-9,3)$
Solution
Let $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=\lambda$
Any point on the line is
$\mathrm{P} \equiv(3 \lambda+6,2 \lambda+7,-2 \lambda+7)$
Given point is $\mathrm{A}(1,2,3)$
$\therefore \quad$ The d.r.s. of the line AP are $3 \lambda+5,2 \lambda+5$, $-2 \lambda+4$
Since the line AP is perpendicular to the given line.
$\therefore \quad 3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4)=0$
$\begin{aligned} & \Rightarrow 17 \lambda+17=0 \\ & \Rightarrow \lambda=-1\end{aligned}$
$\therefore \quad \mathrm{P} \equiv(3,5,9)$