MCQ
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
- A$0$
- B$-1$
- C$-1/2$
- ✓$-2$
✓
Answer
Correct option: D.
$-2$
d
(d) We have, $\frac{{x - 1}}{1} = \frac{{y + 3}}{{ - \lambda }} = \frac{{z - 1}}{\lambda } = s$
(d) 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
$\left| {\,\begin{array}{*{20}{c}}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{l_1}}&{{m_1}}&{{n_1}}\\{{l_2}}&{{m_2}}&{{n_2}}\end{array}\,} \right|\, = \,0$
==> $\left| {\,\begin{array}{*{20}{c}}{ - 1}&4&1\\1&{ - \lambda }&\lambda \\{1/2}&1&{ - 1}\end{array}\,} \right|\, = \,0$
On solving, $\lambda = - 2$.
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