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. $y-2=-1$
• D. $y-2z=-1$

Answer 1

Answer: (B), (C)

PLAN If the straight lines are coplanar. They the should lie in same plane.
Description of Situation If straight lines are coplanar.
$\Rightarrow \left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$
Since, $\frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$
and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar.
$\Rightarrow \left|\begin{array}{ccc}2& 0& 0\\ 2& K& 2\\ 5& 2& K\end{array}\right|=0$
$\Rightarrow$ $K^2=4\Rightarrow K=\pm 2$
$\therefore$ $n_1=b_1\times d_1=6j-6k,fork=2$
$\therefore$ $n_2=b_2\times d_2=14j+14k,fork=-2$
So, equation of planes are $(r-a).n_1=0$
$\Rightarrow$ $y-z=-1$ and $(r-a).n_2=0$
$\Rightarrow$ $y+z=-1$