Question

Find the vector equation of a line passing through point $(1,2,-4)$ and perpendicular to the both 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 required line is
$\overrightarrow{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$...........(1)
Lines $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$
and $\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$
are perpendicular to each other
$3b_1-16b_2+7b_3=0$..................(2)
Similarly DC's of line
$\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$
and $\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$
are $3,8,-5$ perpendicular to each other.
$\therefore 3b_1+8b_2-5b_3=0$...........(3)
$\frac{b_1}{80-56}=\frac{b_2}{21+15}=\frac{b_3}{24+48}$
$\Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}$
Putting the proportional value of $b_1,b_2,b_3$ in (1)
$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$
This is required equation of line.
Cartesian equation of this line is
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$