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, $\frac{1-x}{3}$
$=\frac{y-2}{2\lambda }$
$=\frac{z-3}{2}\dots \left(i\right)$
and $\frac{x-1}{3\lambda }=\frac{y-1}{1}=\frac{6-z}{7}\dots \left(ii\right)$
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
$\left(-3\right)\left(3\lambda \right)+\left(2\lambda \right)\left(1\right)+\left(2\right)\left(-7\right)=0$
$\Rightarrow -9\lambda +2\lambda -14=0$
$\Rightarrow 7\lambda =-14$
$\Rightarrow \lambda =-2$