Question

If lines $x=1+s,y=-3-\lambda s,z=1+\lambda sandx=\frac{t}{2},y=1+t,z=2-t$, where $s$ and $t$ are parameters, are coplanar, then $\lambda$ is equal to:

• A. $-\frac{1}{2}$
• B. $-1$
• C. $-2$
• D. $0$

Answer 1

The correct option is C

$-2$

Explanation for the correct option.

The given straight lines are

$$\begin{gathered} x=1+s,y=-3-\lambda s,z=1+\lambda sand \\ x=\frac{t}{2},y=1+t,z=2-t \end{gathered}$$

where, $s$ and $t$ are parameters. So, the equations will be

$\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }$ and $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{-2}$

Now, if the lines are oplanar then

$\left[\begin{array}{ccc}1-0& -3-1& 1-2\\ 1& -\lambda & \lambda \\ 1& 2& -2\end{array}\right]=0$

Now,

$\begin{array}{rcl}1\left(2\lambda -2\lambda \right)+4\left(-2-\lambda \right)-1\left(2+\lambda \right)& =& 0\\ \Rightarrow -8-4\lambda -2-\lambda & =& 0\\ \Rightarrow 5\lambda & =& -10\\ \Rightarrow \lambda & =& -2\end{array}$

Hence, option C is correct.