Question

Find the value of $\lambda ,$ so that the lines $\frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}\text{and}\frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles. Also, find whether the lines are intersecting or not.

Answer 1

For perpendicular lines, dot product of their d.r.'s is $0.$
$\frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}$
$\Rightarrow \frac{x-1}{-3}=\frac{y-2}{\lambda /7}=\frac{z-3}{2}$
Therefore, d.r.'s of the line $\frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}$ are $(-3,\frac{\lambda }{7},2)$

$\frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5}$
$\Rightarrow \frac{x-1}{-3\lambda /7}=\frac{y-5}{1}=\frac{z-6}{-5}$
Therefore, d.r.'s of the line $\frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5}$ are $(\frac{-3\lambda }{7},1,-5)$

According to the question,
$-3\times \left(\frac{-3\lambda }{7}\right)+\frac{\lambda }{7}\times 1+2\times (-5)=0\Rightarrow \frac{9\lambda }{7}+\frac{\lambda }{7}-10=0\Rightarrow \lambda =7$

Putting the value of $\lambda$ in the equation of the lines we get
$L_1:\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-3}{2}\&L_2:\frac{x-1}{-3}=\frac{y-5}{1}=\frac{z-6}{-5}$
So, any point on the line $L_1$ is of the form $(1-3k_1,k_1+2,2k_1+3)$ and any point on the line $L_2$ is of the form $(1-3k_2,k_2+5,6-5k_2)\text{where}{k}_1,k_2\in R$

If the lines are intersecting, then
$1-3k_1=1-3k_2,k_1+2=k_2+5,2k_1+3=6-5k_2\text{for some}{k}_1,k_2\in R1-3k_1=1-3k_2\Rightarrow k_1=k_2\cdots \left(1\right)k_1+2=k_2+5\Rightarrow k_1=k_2+3\cdots \left(2\right)2k_1+3=6-5k_2\Rightarrow k_1=\frac{3-5k_2}{2}\cdots \left(3\right)$
We can conclude that $\mathrm{\exists ̸}{k}_1,k_2\in R$ which satisfy $\left(1\right),\left(2\right)\text{and}\left(3\right)$ simultaneously.
So, the lines are not intersecting.