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

Question

Let $$\mathrm{N}$$ be the foot of perpendicular from the point $$\mathrm{P}(1,-2,3)$$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2 x-2 y+z+5=0$ 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 $\mathrm{N}(3 \lambda+4,12 \lambda+5,3 \lambda+8)$

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

$$
\overrightarrow{\mathrm{PN}}=(3 \lambda+3) \hat{i}+(12 \lambda+7) \hat{\mathrm{j}}+(3 \lambda+5) \hat{\mathrm{k}}
$$

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

Now, $\overrightarrow{\mathrm{PN}} \cdot \overrightarrow{\mathrm{a}}=0$

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

So point $\mathrm{N}$ is $(2,-3,6)$

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

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

Solutions · 1