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

Answer 1

Answer: (B)

Normal vector of plane containing two intersecting lines is parallel to vector.
$\left({\vec{V}}_1\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& 1\\ -3& 2& -1\end{array}\right|$
$=-2\hat{i}+6\hat{k}$
$\therefore$ Required line is parallel to vector
$\left({\vec{V}}_2\right)=\left(\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ -2& 0& 6\end{array}\right)=3\hat{i}-\hat{j}+\hat{k}$
$\Rightarrow$ Required equation of line is
$\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$