Question

Find the vector and cartesian equations of the line passing through the 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

The line passes through the point $(1,2,-4)$.

The required line is perpendicular to the lines :

$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

Therefore, required line is parallel to :

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|$

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

Hence, required vector equation of line is :

$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (24\hat{i}+36\hat{j}+72\hat{k})$

And, Cartesian equation is :

$\frac{x-1}{24}=\frac{y-2}{36}=\frac{z+4}{72}$