Question

Find the vector equation of the line passing through point $(1,2,-4)$ and perpendicular to the two lines
$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$.

Answer 1

Let the equation of line be $\frac{x-1}{a}=\frac{y-2}{b}=\frac{z+4}{c}$

Now the line is parallel to the two given points so we get

$\left(3\hat{i}-16\hat{j}+7\hat{k}\right)\times \left(3\hat{i}+8\hat{j}-5\hat{k}\right)$

$=24\hat{i}+36\hat{j}+72\hat{k}$

$=2\hat{i}+3\hat{j}+6\hat{k}\left[tak\mathrm{in}g12commom\right]$

Hence,

The required equation is $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$