Find the vector equation of the line passing through the point (2,3,-1) and parallel to the planes <img alt="\vec{r}\cdot \left ( 3\hat{i}+4\hat{j}+2\hat{k} \right )= 5" src="https://entrancecorner.codecogs.com/gif.latex?%5Cvec%7Br%7D%5Ccdot%20%5Cleft%

Find the vector equation of the line passing through the point (2,3,-1) and parallel to the planes $\vec{r}\cdot \left ( 3\hat{i}+4\hat{j}+2\hat{k} \right )= 5$ and $\vec{r}\cdot \left ( 3\hat{i}-2\hat{j}-2\hat{k} \right )= 4$ .

Answers (1)

Suppose the required line is parallel to the vector $\vec{b}$ which is given by $\vec{b}= b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$
The position vector of the points is (2,3,-1) is $\vec{a}= 2\hat{i}+3\hat{j}-\hat{k}\left$
The equation of the line passing through (2,3,-1) & parallel to $\vec{b}$ is given as $\vec{r}= \vec{a}+\lambda \vec{b}$
$\vec{r}=\left ( 2\hat{i}+3\hat{j}-\hat{k} \right )+\lambda \left ( b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k} \right )---(1)$
The equations of the given plane are
$\vec{r}\cdot \left ( 3\hat{i}+4\hat{j}+2\hat{k} \right )= 5---(2)$
&
$\vec{r}\cdot \left ( 3\hat{i}-2\hat{j}-2\hat{k} \right )= 4---(3)$
The line in equation (1) & plane in equation (2) are parallel
$\therefore$ The normal to plane in equation (2) and the given line are perpendiculars
$\therefore = \left ( 3\hat{i}+4\hat{j}+2\hat{k} \right )\cdot \lambda \left ( b_{1}\hat{i}+b_{2}\hat{j}+b_{3} \hat{k}\right )= 0$
$\Rightarrow \lambda \left ( 3b_{1}+4b_{2}+2b_{3} \right )= 0$
$\Rightarrow 3b_{1}+4b_{2}+2b_{3}= 0---(4)$
similarly, from equation (1) & (3) we get
$\left ( 3\hat{i}-2\hat{j}-2\hat{k} \right )\lambda \left ( b_{1}\hat{i}+ b_{2}\hat{j}+ b_{3}\hat{k} \right )= 0$
$\Rightarrow \lambda \left ( 3b_{1}-2b_{2}-2b_{3} \right )= 0$
$\Rightarrow 3b_{1}-2b_{2}-2b_{3} = 0---(5)$
from equation (4) & (5) we get
$\frac{b_{1}}{4\left ( -2 \right )-\left ( -2 \right )\left ( 2 \right )}= \frac{b_{2}}{3\left ( -2 \right )-3\left ( 2 \right )}= \frac{b_{3}}{3\left ( -2 \right )-3\times 4}$
$\frac{b_{1}}{-4}= \frac{b_{2}}{-12}= \frac{b_{3}}{-18}$
$\therefore$ The direction ratios of $\vec{b}$ are -4,-12,-18
$\therefore \vec{b}= -4\hat{i}-12\hat{j}-18\hat{k}\: \left [ \because \vec{b}= b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k} \right ]$
On substituting the value of $\vec{b}$ in equation (1) we get
$\vec{r}= \left ( 2\hat{i} +3\hat{j}-\hat{k}\right )+\lambda \left (-4\hat{i} -12\hat{j}-18\hat{k} \right )$ which is the required solution and required equation of the line.