Three Dimensional Geometry Question 68
Question: If the straight lines $ x=1+s, $ $ y=-3-\lambda s, $ $ z=1+\lambda s $ and $ x=t/2,y=1+t,z=2-t $ , with parameters s and $ t $ respectively, are co-planar, then $ \lambda $ equals
[AIEEE 2004]
Options:
A) 0
B) -1
C) -1/2
D) -2
Show Answer
Answer:
Correct Answer: D
Solution:
We have, $ \frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s $
and $ \frac{x-0}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}=t $
Since, lines are coplanar then
$ | ,\begin{matrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ \end{matrix}, |,=,0 $
Þ $ | ,\begin{matrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \\ \end{matrix}, |,=,0 $
On solving, $ \lambda =-2 $ .
BETA
