Let P be the point (10,-2,-1) and Q be the foot of the perpendicular drawn from the point …

Question

Let $P$ be the point $(10,-2,-1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1,7,6)$ on the line passing through the points $(2,-5,11)$ and $(-6,7,-5)$. Then the length of the line segment $P Q$ is equal ...

Answer: 13

Accepted Answer

$\begin{gathered} P(10,-2,-1) \ MN:\frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16} \end{gathered}$

General point

$(8k+2,-12k-5,16k+11)$

$\overset{	o \leftarrow }{RQ}=(8k+2-1)\hat{i}+(-12k-5-7)\hat{j}+(16k+11-6)\hat{k}$

$\overset{	o \leftarrow }{RQ}=(8k+1)\hat{i}-(12k+12)\hat{j}+(16k+5)\hat{k}$

$\overset{	o \leftarrow }{RQ}\, .\, \overset{	o \leftarrow }{MN}=0$ (as both are perpendicular)

$8(8k+1)+12(12k+12)+16(16k+5)=0$

$\begin{gathered} 64k+8+144k+144+256k+80=0 \ 464k=-232 \ k=\frac{-232}{464}=\frac{-1}{2} \ Q(-4+2,6-5,-8+11) \ Q(-2,1,3) \ PQ=\sqrt{(10+2{)}^{2}+(-3{)}^{2}+(4{)}^{2}} \ PQ=\sqrt{{12}^{2}+{3}^{2}+{4}^{2}} \ PQ=\sqrt{169}=13 \end{gathered}$

Solutions · 1