Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the lines $\displaystyle \vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})$ and $\displaystyle \vec{r}=(3 \hat{i}+3 \hat{j})+\mu(2 \hat{i}+\hat{j}+\hat{k})$ respectively.

Based on the above information, answer the following questions :

(a)

Find the shortest distance between the given lines.

(b)

Find the point at which the motorcycles may collide. $\displaystyle 7$

\[\begin{gathered} \vec{a}_{1}=0 \hat{i}+0 \hat{j}+0 \hat{k}, \vec{a}_{2}=3 \hat{i}+3 \hat{j} \\ \vec{a}_{2}-\vec{a}_{1}=3 \hat{i}+3 \hat{j} \\ \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 1 \end{array}\right| \\ =3 \hat{i}-3 \hat{j}-3 \hat{k} \\ \mathrm{SD}=\frac{\left|\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)\right|}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|} \end{gathered} \] Now, \[\begin{aligned} \left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right) & =(3 \hat{i}+3 \hat{j})(3 \hat{i}-3 \hat{j}-3 \hat{k}) \\ & =9-9=0 \end{aligned} \] Shortest distance between two lines \(\displaystyle =0\) Any point on the line \(\displaystyle \vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})\) is \(\displaystyle \lambda \hat{i}+2 \lambda \hat{j}-\lambda \hat{k}\) As the lines are intersecting, As the lines are intersecting,