Question

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

Answer 1

$L_1:\frac{x+3}{3}=\frac{y-4}{5}=\frac{x+8}{5}$

$\overrightarrow{Q}=-3\hat{i}+4\hat{j}-8\hat{z}$ $Q\equiv (-3,4,-8)$

$\overrightarrow{b}=3\hat{i}+5\hat{j}+5\hat{k}$

$P(-2,4,-5)$

$\overrightarrow{PQ}=(-3\hat{i}+4\hat{j}-8\hat{z})-(-2\hat{i}+4\hat{j}-5\hat{k})=-\hat{i}-3\hat{k}$

$\overrightarrow{b}\times \overrightarrow{PQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 5\\ -1& 0& -3\end{array}\right|=\hat{i}\left|\begin{array}{cc}5& 5\\ 0& -3\end{array}\right|-\hat{j}\left|\begin{array}{cc}3& 5\\ -1& -3\end{array}\right|+\hat{k}\left|\begin{array}{cc}3& 5\\ -1& 0\end{array}\right|$

$=-15\hat{i}-\hat{j}[-9+5]+\hat{k}\left[5\right]$

$=-15\hat{i}+4\hat{j}+5\hat{k}$

$|\overrightarrow{b}\times \overrightarrow{PQ}|=\sqrt{225+16+125}=\sqrt{266}$

Hence, $d=\frac{|\overrightarrow{b}\times \overrightarrow{PQ}|}{|\overrightarrow{b}|}=\frac{\sqrt{266}}{\sqrt{9+25+25}}=\frac{\sqrt{266}}{\sqrt{59}}=\sqrt{\frac{266}{59}}$ units.