Question

If the lines $\frac{1-x}{3}=\frac{y-2}{2\alpha }=\frac{z-3}{2}$ and $\frac{x-1}{3\alpha }$ $=y-1=\frac{6-z}{5}$ are perpendicular, then the value of $\alpha$ is

• A. $\frac{-10}{7}$
• B. $\frac{10}{7}$
• C. $\frac{-10}{11}$
• D. $\frac{10}{11}$
• E. $\frac{10}{9}$

Answer 1

Answer: (A)

Given lines can be rewritten as $\frac{x-1}{-3}=\frac{y-2}{2\alpha }=\frac{z-3}{2}$
and $\frac{x-1}{3\alpha }=\frac{y-1}{1}=\frac{z-6}{-5}$
Since, lines are perpendicular.
$\therefore$ $a_1{a}_2+b{}_1{b}_2+c_1{c}_2=0$
$\Rightarrow$ $(-3)(3\alpha )+2\alpha (1)+(2)-5=0$
$\Rightarrow$ $-9\alpha +2\alpha -10=0$
$\Rightarrow$ $\alpha =-\frac{10}{7}$