The lines x = ay $-$ 1 = z $-$ 2 and x = 3y $-$ 2 = bz $-$ 2, (ab $\ne$ 0) are coplanar, if :
Solution
Lines are $x = ay - 1 = z - 2$<br><br>$\therefore$ ${x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}$ .... (i)<br><br>and $x = 3y - 2 = bz - 2$<br><br>$\therefore$ $${x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}$$ .... (ii)<br><br>$\therefore$ lines are co-planar<br><br>$\therefore$ $$\left| {\matrix{
0 & { - {1 \over a} + {2 \over 3}} & { - 2 + {2 \over b}} \cr
1 & {{1 \over a}} & 1 \cr
1 & {{1 \over 3}} & {{1 \over b}} \cr
} } \right| = 0$$<br><br>$\therefore$ $$\left| {\matrix{
0 & {{2 \over 3} - {1 \over a}} & {{2 \over b} - 2} \cr
0 & {{1 \over a} - {1 \over 3}} & {1 - {1 \over b}} \cr
1 & {{1 \over 3}} & {{1 \over b}} \cr
} } \right| = 0$$<br><br>$\therefore$ ${1 \over a} - {1 \over {ab}} = 0$<br><br>$\Rightarrow$ b = 1 and a $\in$ R $-$ {0}
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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