Question

If the lines $x=1+a,y=-3-\lambda a,z=1+\lambda a$ and $x=\frac{b}{2},y=1+b,z=2-b$ are coplanar, then $\lambda$ is equal to

• A. $-3$
• B. $2$
• C. $1$
• D. $-2$

Answer 1

The correct option is D $-2$

The given lines are $\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=\left(a\right)$ and $\frac{x-0}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}=\left(b\right)$

$\therefore$ coplanarity, we must have

$\left|\begin{array}{ccc}-1& 4& 1\\ 1& -\lambda & \lambda \\ 1/2& 1& -1\end{array}\right|=0$

$\Rightarrow -1(\lambda -\lambda )-4(-1-\frac{\lambda }{2})+(1+\frac{\lambda }{2})=0$

$4+2\lambda +1+\frac{\lambda }{2}=0\Rightarrow 5+\frac{5\lambda }{2}=0\therefore \lambda =-2$