Question

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

Answer 1

Let the required line be parallel to vector $\overrightarrow{b}$ is given by,
$\overrightarrow{r}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$

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

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

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

$\Rightarrow$ $\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})).....(1)$

The equations of the given planes are

$\overrightarrow{r}.(\hat{i}-\hat{j}+2\hat{k})=\mathrm{5.......}\left(2\right)$

$\overrightarrow{r}.(3\hat{i}+\hat{j}+\hat{k})=\mathrm{6.......}\left(3\right)$

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.

$\Rightarrow$ $(\hat{i}-\hat{j}+2\hat{k}).\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k}))=0$

$\Rightarrow$ $\lambda (b_1-b_2+2b_3)=0$

$\Rightarrow$ $b_1-b_2+2b_3=\mathrm{0.......}\left(4\right)$

Similarly, $(3\hat{i}+\hat{j}+\hat{k}).\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k}))=0$

$\Rightarrow$ $\lambda (3b_1+b_2+b_3)=0$

$\Rightarrow$ $3b_1+b_2+b_3=\mathrm{0........}\left(5\right)$

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 $\overrightarrow{b}$ are $-3,5$ and $4$.

$\therefore$ $\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=-3\hat{i}+5\hat{j}+4\hat{k}$

Substituting the value of $\overrightarrow{b}$ in equation (1), we obtain
$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (-3\hat{i}+5\hat{j}+4\hat{k})$

This is the equation of the required line.