Question

The value of $ \lambda $ for which the lines
$ \frac{1-x}{3} = \frac{y-2}{2\lambda}=\frac {z-3}{2} = \frac{x-1}{3\lambda} =\frac{y-1}{1} =\frac {6-z}{7} $
are perpendicular to each other is

• A. -1
• B. -2
• C. 1
• D. 2

Answer 1

Answer: (B)

Given, $\frac{1-x}{3}$
$=\frac{y-2}{2\lambda}$
$=\frac{z-3}{2} \ldots\left(i\right)$
and $\frac{x-1}{3\lambda} =\frac{y-1}{1} =\frac{6-z}{7} \ldots\left(ii\right)$
Eqs.$ \left(i\right)$ and $\left(ii\right)$, can be written as
$\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}$
and $ \frac{x-1}{3\lambda} =\frac{y-1}{1}$
$=\frac{z-6}{-7}$
Since, both are perpendicular to each other, then
$\left(-3\right)\left(3\lambda\right)+\left(2\lambda\right)\left(1\right)+\left(2\right)\left(-7\right)=0$
$\Rightarrow -9\lambda+2\lambda-14=0$
$\Rightarrow 7\lambda=-14$
$\Rightarrow \lambda=-2$