Question

If the straight lines $\frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{K}$ are coplanar, then the plane(s) containing these two lines is(are)

• A. $y+2z=-1$
• B. $y+z=-1$
• C. $y-z=-1$
• D. $y-2z=-1$

Answer 1

Answer: (B)

$y+z=-1$

Two straight lines
$\frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{K}$ are co-planar, then we get

$\left|\begin{array}{ccc}2& K& 2\\ 5& 2& K\\ 2& 0& 0\end{array}\right|=0$

$\Rightarrow K^2=4\Rightarrow K=\pm 2$

For $K=2$, the plane $y+1=z$ is common in both lines.

For $K=-2$, family of plane containing first line is $x+y+\lambda (x-z-1)=0$

Point $(-1,-1,0)$ must satisfy $\lambda$.

$\Rightarrow -2+\lambda (-2)=0\Rightarrow \lambda =-1$

$\Rightarrow y+z+1=0$
$\Rightarrow y+z=-1$