Mathematics - Vector Question with Solution | TestHub

Given the lines $L_1:\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}$ and $L_2:\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$.

From the given lines, we can say that both the lines pass through the point $\left(1, 0, 1\right)$.

Now, let $\overrightarrow{a}$ be the direction vector of line $L_1$ i.e., $\overrightarrow{a}=i-j+3k$ and let $\overrightarrow{b}$ be the direction vector of line $L_2$ i.e., $\overrightarrow{b}=-3i-j+k$.

We also know that the angle bisectors lie in the direction of $\left(\overrightarrow{a}+\overrightarrow{b}\right) \text{or} \left(\overrightarrow{a}-\overrightarrow{b}\right)$.

Now, $\overrightarrow{a}.\overrightarrow{b}=\left(1\times -3\right)+\left(-1\times -1\right)+\left(3\times 1\right)=-3+1+3=1>0$.

So, direction ratios of the acute angle bisector between two lines will be in the direction of $\overrightarrow{a}+\overrightarrow{b}$ i.e., $\left(1-3, -1-1, 3+1\right)=\left(-2 , -2 , 4\right)=\left(1, 1, -2\right)$.

Since, $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$ is the acute angle bisector between the lines $L_1 \& L_2$.

So, by comparing the direction ratios, we get

$l=1 \& m=1$

$\Rightarrow l+m=2$

So, equation of line $L$ will be $\frac{x-\alpha }{1}=\frac{y-1}{1}=\frac{z-\gamma }{-2}$.

Given that the line $L$ passes through the point $\left(1, 0, 1\right)$.

Putting the point in line $L$, we get 

$\frac{1-\alpha }{1}=\frac{0-1}{1}=\frac{1-\gamma }{-2}$

$\Rightarrow 1-\alpha =-1=\frac{1-\gamma }{-2}$

$\Rightarrow 1-\alpha =-1 \& \frac{1-\gamma }{-2}=-1$

$\Rightarrow \alpha =2 \& \gamma =-1$

$\Rightarrow \alpha -\gamma =3$

So, options $\left(A\right) \& \left(B\right)$ are correct.