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} $