Question

The equation of the line passing through $(-4,3,1)$, parallel to the plane $x+2y-z-5=0$ and intersecting the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$ is:

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

Answer 1

The correct option is C $\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$

Vector passing through $(-4,3,1)$ and $(-1,3,2)=3\hat{i}+\hat{k}$
Normal vector of plane containing two intersecting lines is parallel to the vector.
$\overrightarrow{r_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& 1\\ -3& 2& -1\end{array}\right|=-2\hat{i}+6\hat{k}$

$\therefore$ Required line is parallel to the vector.
$\overrightarrow{r_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ -2& 0& 6\end{array}\right|=3\hat{i}-\hat{j}+\hat{k}$
$\therefore$ Required equation of line passing through $(-4,3,1)$ is $\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$