Question

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

Answer 1

$\begin{array}{c}Givenline:D.R\Rightarrow (3,5,6)andrequiredpoints\Rightarrow (-2,4,-5)\\ linevectoerequ:\\ \overrightarrow{r}=-2\hat{i}+4\hat{j}-5\hat{k}+\lambda (3\hat{i}+5\hat{j}+6\hat{k})\\ Now,\\ InCartesianformpointwithD.R,\\ \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\\ \end{array}$