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 straight lines are coplanar then
$\Rightarrow \left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

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 $(r-a).n_1=0$
$\Rightarrow y-z=-1and(r-a).n_2=0\Rightarrow y+z=-1$