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), (C)

$y-z=-1$

If two lines are coplanar then
$\left|\begin{array}{ccc}2& 0& 0\\ 2& k& 2\\ 5& 2& k\end{array}\right|=0$
$\Rightarrow k^2-4=0\Rightarrow k=\pm 2$
So lines are
$\frac{x-1}{2}=\frac{y+1}{2}=\frac{z}{2}$
and
$\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{2}$
OR
$\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{-2}$
Now equation of plane through these lines will be
$\left|\begin{array}{ccc}x-1& y+1& z\\ 1& 1& 1\\ 5& 2& 2\end{array}\right|=0$ (or) $\left|\begin{array}{cccc}x-1& y+1& z& \\ 1& -1& 1& \\ 5& 2& -2& \end{array}\right|=0$
i.e $y-z+1=0\text{or}y+z+1=0$