# Find the vector equation of the line passing through the point

Question:

Find the vector equation 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 $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

Solution:

Let the required line be parallel to the vector $\vec{b}$ given by, $\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$

The position vector of the point $(1,2,-4)$ is $\vec{a}=\hat{i}+2 \hat{j}-4 \hat{k}$

The equation of the line passing through $(1,2,-4)$ and parallel to vector $\vec{b}$ is

$\vec{r}=\vec{a}+\lambda \vec{b}$

$\Rightarrow \vec{r}(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)$                       $\ldots(1)$

The equations of the lines are

$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$                                                                                                                                    $\ldots(2)$

$\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$                                                                                                                                       $\ldots(3)$

Line (1) and line (2) are perpendicular to each other.

$\therefore 3 b_{1}-16 b_{2}+7 b_{3}=0$                                                                                                                                             $\ldots(4)$

Also, line (1) and line (3) are perpendicular to each other.

$\therefore 3 b_{1}+8 b_{2}-5 b_{3}=0$                                                                                                                                               $\ldots(5)$

From equations (4) and (5), we obtain

$\frac{b_{1}}{(-16)(-5)-8 \times 7}=\frac{b_{2}}{7 \times 3-3(-5)}=\frac{b_{3}}{3 \times 8-3(-16)}$

$\Rightarrow \frac{b_{1}}{24}=\frac{b_{2}}{36}=\frac{b_{3}}{72}$

$\Rightarrow \frac{b_{1}}{2}=\frac{b_{2}}{3}=\frac{b_{3}}{6}$

$\therefore$ Direction ratios of $\vec{b}$ are 2,3, and 6 .

$\therefore \vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}$

Substituting $\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}$ in equation (1), we obtain

$\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$

This is the equation of the required line.