Question

The lines $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$
and $\frac{x-4}{1}=\frac{y+3}{k}=\frac{z+1}{7}$ are coplanar if $k =$

• A. $4$
• B. $-4$
• C. $2$
• D. $-2$

Answer 1

Answer: (B)

Given lines are $\frac{x-1}{2}=\frac{y-\left(-1\right)}{-3}=\frac{z-\left(-10\right)}{8}$
and $\frac{x-4}{1}=\frac{y-\left(-3\right)}{k}=\frac{z-\left(-1\right)}{7}$
These lines are coplanar
$\therefore \begin{vmatrix}4-1&-3-\left(-1\right)&-1-\left(-10\right)\\ 2&-3&8\\ 1&k&7\end{vmatrix}=0$
$\Rightarrow 3\left(- 21 - 8k\right) + 2\left(14 - 8\right) + 9\left(2k + 3\right) = 0$
$\Rightarrow 6k=-24$
$\therefore k=-4$.