If the lines $$\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$$ and $$\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$$ are co-planar, then the distance of the plane containing these two lines from the point ($\alpha$, 0, 0) is :
Solution
<p>$\because$ Both lines are coplanar, so</p>
<p>$$\left| {\matrix{
{\alpha - 1} & 0 & { - 1} \cr
0 & 3 & { - 1} \cr
2 & 0 & { - 3} \cr
} } \right| = 0$$</p>
<p>$\Rightarrow \alpha = {5 \over 3}$</p>
<p>Equation of plane containing both lines</p>
<p>$$\left| {\matrix{
{x - 1} & {y + 1} & {z - 1} \cr
0 & 3 & { - 1} \cr
2 & 0 & { - 3} \cr
} } \right| = 0$$</p>
<p>$\Rightarrow 9x + 2y + 6z = 13$</p>
<p>So, distance of $\left( {{5 \over 3},0,0} \right)$ from this plane</p>
<p>$= {2 \over {\sqrt {81 + 4 + 36} }} = {2 \over {11}}$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
This question is part of PrepWiser's free JEE Main question bank. 182 more solved questions on Three Dimensional Geometry are available — start with the harder ones if your accuracy is >70%.