Question

The equation of the line passing through $(1,2,3)$ and parallel to the planes
$\overline{r}\cdot (\overline{i}-\overline{j}+2\overline{k})=5$ and $\overline{r}\cdot (3\overline{i}+\overline{j}+\overline{k})=6$ is ?

• A. $\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}$
• B. $\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$
• C. $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{4}$
• D. $\frac{x-1}{5}=\frac{y-2}{4}=\frac{z-3}{3}$

Answer 1

The correct option is B $\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$

Equation of line passing through a point with position vector $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$

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

Point: $(1,2,3)$

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

Given that the line is parallel to both the planes i.e. it is perpendicular to normal of both the planes.

$\therefore \overrightarrow{b}=$ cross product of the direction vectors of the given planes $\overrightarrow{r}\cdot (\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k})=5$ and $\overrightarrow{r}\cdot (3\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})=6$

$\overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 3& 1& 1\end{array}\right|=-3\overrightarrow{i}+5\overrightarrow{j}+4\overrightarrow{k}$

The required equation is

$\overrightarrow{r}=\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k}+\lambda \left(-3\overrightarrow{i}+5\overrightarrow{j}+4\overrightarrow{k}\right)$

$\frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4}$