Let N be the foot of perpendicular from the point P(1,-2,3) on the line passing through th…

Question

Let $N$ be the foot of perpendicular from the point $P(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2x-2y+z+5=0$ is :

Accepted Answer

Equation of line
$\frac{x-4}{4-1}=\frac{y-5}{5-(-7)}=\frac{z-8}{8-5}$

$\frac{x-4}{3}=\frac{y-5}{12}=\frac{z-8}{3}$

Let point $N(3\lambda +4,12\lambda +5,3\lambda +8)$

$\overset{	o \leftarrow }{PN}=(3\lambda +4-1)\hat{i}+(12\lambda +5-(-2))\hat{j}+(3\lambda +8-3)\hat{k}$

$\overset{	o \leftarrow }{PN}=(3\lambda +3)\hat{i}+(12\lambda +7)\hat{j}+(3\lambda +5)\hat{k}$

And parallel vector to line (say $\vec{a}=3\hat{i}+12\hat{j}+3\hat{k}$ )

Now, $\overset{	o \leftarrow }{PN}\cdot \overset{	o }{a}=0$

$\begin{gathered} (3\lambda +3)3+(12\lambda +7)12+(3\lambda +5)3=0 \ 162\lambda +108=0\Rightarrow \lambda =\frac{-108}{162}=\frac{-2}{3} \end{gathered}$

So point $N$ is $(2,-3,6)$

Now distance of
N(2, -3, 6) from
$2x-2y+z+5=0$ is

$=|\frac{2(2)-2(-3)+6+5}{\sqrt{4+4+1}}|=7$

Solutions · 1