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

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

Let the point of intersection be $(-3\lambda -1,2\lambda +3,-\lambda +2)$

Direction cosines of line is $(-3\lambda -1(-4),2\lambda +3-3,-\lambda +2-1)=(-3\lambda +3,2\lambda ,-\lambda +1)$

Since, the line is parallel to $x+2y-z=5$. Hence dot product is zero.

$(-3\lambda +3)1+\left(2\lambda \right)2+(-\lambda +1)(-1)=0$

$\Rightarrow -3\lambda +3+4\lambda +\lambda -1=0$

$\Rightarrow 2\lambda +2=0$

$\Rightarrow \lambda =-1$

Hence direction cosines are $(-3(1)+3,2(-1),-(-1)+1)=(6,-2,2)=2(3,-1,1)$

Thus equation of the line is $\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$.