Question

The distance of the point $(-2,4,-5)$ from the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is

• A. $\frac{\sqrt{37}}{10}$
• B. $\sqrt{\frac{37}{10}}$
• C. $\frac{37}{\sqrt{10}}$
• D. $\frac{37}{10}$

Answer 1

Answer: (B)

The line passes through $A(-3,4,-8)$ and is parallel to the vector $b=3\hat{i}+5\hat{j}+6\hat{k}$

Let $M$ be the foot of the perpendicular from $P(-2,4,-5)$ on the given line.
We, have
$AP=-\hat{i}+0\hat{j}-3\hat{k}$
$\Rightarrow |AP|=\sqrt{1+0+9}=\sqrt{10}$
Clearly, AM = Projection of AP on $b$
$\Rightarrow AM=\left|\frac{AP\cdot b}{b}\right|$
$=\left|\frac{\mid -\hat{i}-3\hat{k})\cdot (3\hat{i}+5\hat{j}+6\hat{k})}{|3\hat{i}+5\hat{j}+6\hat{k}|}\right|$
$=\left|\frac{-3-18}{\sqrt{9+25+36}}\right|=\left|\frac{-21}{\sqrt{70}}\right|=\frac{21}{\sqrt{70}}$
$\therefore PM=\sqrt{A{P}^2-A{M}^2}$
$=\sqrt{10-\frac{441}{70}}=\sqrt{\frac{259}{70}}$
$=\sqrt{\frac{37}{10}}$