If for some $\alpha$ $\in$ R, the lines
L1 : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and
L2 : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar,
then the line L2
passes through the point :
Solution
L<sub>1</sub> : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and
<br><br>L<sub>2</sub> : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar.
<br><br>$\therefore$ $$\left| {\matrix{
1 & 3 & 2 \cr
2 & { - 1} & 1 \cr
\alpha & {5 - \alpha } & 1 \cr
} } \right|$$ = 0
<br><br>$\Rightarrow$ –1(–1 + $\alpha$ - 5) + 3(2 - $\alpha$) - 2(10 - 2$\alpha$ + $\alpha$) = 0
<br><br>$\Rightarrow$ 6 - $\alpha$ + 6 - 3$\alpha$ + 2$\alpha$ - 20 = 0
<br><br>$\Rightarrow$ –8 –2$\alpha$ = 0
<br><br>$\Rightarrow$ $\alpha$ = -4
<br><br>$\therefore$ Equation of L<sub>2</sub> : ${{x + 2} \over { - 4}} = {{y + 1} \over 9} = {{z + 1} \over 1}$
<br><br>Check options (2, –10, –2) lies on L<sub>2</sub>.
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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