Question

The lines $x=ay–1=z–2$ and $x=3y–2=bz–2,$ $(ab\ne 0)$ are coplanar, if

• A. $a=2,b=2$
• B. $a=2,b=3$
• C. $a=1,b\in R-\left\{0\right\}$
• D. $b=1,a\in R-\left\{0\right\}$

Answer 1

The correct option is D $b=1,a\in R-\left\{0\right\}$

Lines are $x=ay–1=z–2$
$\therefore \frac{x}{1}=\frac{y-\frac{1}{a}}{\frac{1}{a}}=\frac{z-2}{1}...\left(i\right)$
and $x=3y–2=bz–2$
$\therefore \frac{x}{1}=\frac{y-\frac{2}{3}}{\frac{1}{3}}=\frac{z-\frac{2}{b}}{\frac{1}{b}}...\left(ii\right)$
Since, lines are co-planar.
$\therefore \left|\begin{array}{ccc}0& -\frac{1}{a}+\frac{2}{3}& -2+\frac{2}{b}\\ 1& \frac{1}{a}& 1\\ 1& \frac{1}{3}& \frac{1}{b}\end{array}\right|=0$

$\therefore \left|\begin{array}{ccc}0& \frac{2}{3}-\frac{1}{a}& \frac{2}{b}-2\\ 0& \frac{1}{a}-\frac{1}{3}& 1-\frac{1}{b}\\ 1& \frac{1}{3}& \frac{1}{b}\end{array}\right|=0$
$\therefore \frac{1}{a}-\frac{1}{ab}=0$
$\Rightarrow b=1$ and $a\in R-\left\{0\right\}$