Mathematics - Vector Question with Solution | TestHub

Let point on $L_1$ be $P\left(\left(1+\lambda \right)\hat{i}+\left(2-\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k}\right)$ and point on $L_2$ be $Q\left(\left(4+\mu \right)\hat{i}+\left(5+\mu \right)\hat{j}+\left(6-\mu \right)\hat{k}\right)$.

$\Rightarrow \overrightarrow{PQ}=\left(\lambda -\mu -3\right)\hat{i}+\left(-\lambda -\mu -3\right)\hat{j}+\left(\lambda +\mu -3\right)\hat{k}$

Now, $\overrightarrow{PQ} \perp L_1, L_2$

$\Rightarrow \left(\lambda -\mu -3\right)+\left(\lambda +\mu +3\right)+\left(\lambda +\mu -3\right)=0$

$\Rightarrow 3\lambda +\mu =3 ...\left(i\right)$

Also, $\left(\lambda -\mu -3\right)+\left(-\lambda -\mu -3\right)+\left(-\lambda -\mu +3\right)=0$

$\Rightarrow -\lambda -3\mu =3 ...\left(ii\right)$

Now, solving above equations we get,

$\Rightarrow 3\lambda +\mu -3\lambda -9\mu =3+9$

$\Rightarrow -8\mu =12$

$\Rightarrow \mu =\frac{-3}{2}$

$\Rightarrow \lambda =\frac{3}{2}$

$\Rightarrow P\equiv \left(\frac{5}{2},\frac{1}{2},\frac{9}{2}\right), Q\equiv \left(\frac{5}{2},\frac{7}{2},\frac{15}{2}\right)$

So, mid-point of $PQ$ is $\left(\frac{5}{2},2,6\right)$

$\Rightarrow 2\left(\alpha +\beta +\gamma \right)=2\left(\frac{5}{2}+2+6\right)$

$\Rightarrow 2\left(\alpha +\beta +\gamma \right)=21$