Question

Find the equations of the straight line passing through the point $(1,2,3)$ to intersect the straight line $x+1=2(y-2)=z+4$ and parallel to the plane $x+5y+4z=0$

• A. $\frac{x-1}{2}=\frac{y-2}{2}=\frac{z-3}{-3}$
• B. $\frac{x-1}{2}=\frac{y-3}{3}=\frac{z-3}{-3}$
• C. $\frac{x-1}{2}=\frac{y-2}{2}=\frac{z-3}{3}$
• D. None of these

Answer 1

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

Let $a,b,c$ are $D{r}^{\prime }s$ of required line, thus equation of line passing through $(1,2,3)$

$\frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}=k$(say) ...(1)

$\Rightarrow (ak+1,bk+2,ck+3)$ is any general point on (1).

Also, given line that intersect (1) is

$\frac{x-(-1)}{2}=\frac{y-2}{1}=\frac{z-(-4)}{2}=\lambda$ ....(2)

$\Rightarrow (2\lambda -1,\lambda +2,2\lambda -4)$ is any general point on (2)

$\because$ line (1) and (2) are intersecting

$a=\frac{2\lambda -2}{k};b=\frac{\lambda }{k};c=\frac{2\lambda -7}{k}$

$\because$ (1) is parallel to plane $x+5y+4z=0$ i.e. perpendicular to normal vector.

So $a+5b+4c=0$

$\Rightarrow \left(\frac{2\lambda -2}{k}\right)+5\frac{\lambda }{k}+4\left(\frac{2\lambda -7}{k}\right)=0\Rightarrow \lambda =2$

$\Rightarrow a=\frac{2}{k};b=\frac{2}{k};c=-\frac{3}{k}$

Therefore required line is $\frac{x-1}{2}=\frac{y-2}{2}=\frac{z-3}{-3}$