# Find the vector equation of the line passing through $(1,2,3)$ and parallel to the planes

Question:

Find the vector equation of the line passing through $(1,2,3)$ and parallel to the planes $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6$.

Solution:

Let the required line be parallel to vector given by,

$\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$

The position vector of the point $(1,2,3)$ is $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$

The equation of line passing through $(1,2,3)$ and parallel to $\vec{b}$ is given by,

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

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

The equations of the given planes are

$\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5$                                                       $\ldots(2)$

$\vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6$                                                      $\ldots(3)$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

$\Rightarrow(\hat{i}-\hat{j}+2 \hat{k}) \cdot \lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)=0$

$\Rightarrow \lambda\left(b_{1}-b_{2}+2 b_{3}\right)=0$

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

Similarly, $(3 \hat{i}+\hat{j}+\hat{k}) \cdot \lambda\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)=0$

$\Rightarrow \lambda\left(3 b_{1}+b_{2}+b_{3}\right)=0$

$\Rightarrow 3 b_{1}+b_{2}+b_{3}=0$                                                             $\ldots(5)$

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

$\frac{b_{1}}{(-1) \times 1-1 \times 2}=\frac{b_{2}}{2 \times 3-1 \times 1}=\frac{b_{3}}{1 \times 1-3(-1)}$

$\Rightarrow \frac{b_{1}}{-3}=\frac{b_{2}}{5}=\frac{b_{3}}{4}$

Therefore, the direction ratios of $\vec{b}$ are $-3,5$, and $4 .$

$\therefore \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}=-3 \hat{i}+5 \hat{j}+4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})$

This is the equation of the required line.