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.
$\Rightarrow \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ \end{vmatrix}= 0$
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.
$\Rightarrow \begin{vmatrix} 2 & 0 & 0 \\ 2 & K & 2 \\ 5 & 2 & K \\ \end{vmatrix}= 0$
$\Rightarrow $ $ K^2 = 4 \Rightarrow K = \pm 2 $
$\therefore $ $ n_1 = b_1 \times d_1 = 6j - 6k, for\ k = 2$
$\therefore $ $ n_2 = b_2 \times d_2 = 14j + 14k, for\ k = -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$