Mathematics - Vector Question with Solution | TestHub

Given: $L_1\perp L_2$

$\Rightarrow \left(\hat{i}-\hat{j}+2\hat{k}\right).\left(3\hat{i}+\hat{j}+p\hat{k}\right)=0$

$\Rightarrow 3-1+2p=0$

$\Rightarrow p=-1$

Also, $L_3\perp L_1, L_2$

So, $L_3\parallel \left(L_1\times L_2\right)$

$\Rightarrow L_1\times L_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 3& 1& -1\end{array}\right|$

$\Rightarrow L_1\times L_2=-\hat{i}+7\hat{j}+4\hat{k}$

On comparing with $L_3:\overrightarrow{r}=\delta (l\hat{i}+m\hat{j}+n\hat{k})$, we get that $(-\delta ,7\delta ,4\delta )$ will lie on $L_3$.

Now, for $\delta =1$ the point will be $(-1,7,4)$.