(a)OR(b)

Verify that lines given by $\displaystyle \vec{r}=(1-\lambda) \hat{\mathrm{i}}+(\lambda-2) \hat{\mathrm{j}}+(3-2 \lambda) \hat{k}$ and $\displaystyle \vec{r}=(\mu+1) \hat{\mathrm{i}}+(2 \mu-1) \hat{\mathrm{j}}-(2 \mu+1) \hat{k}$ are skew lines. Hence, find shortest distance between the lines.

Rewriting the lines, we get \[\vec{r}=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{k})+\lambda(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{k}) \text { and } \vec{r}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{k})+\mu(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{k}) \] Let $\displaystyle \vec{a}_{1}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{k}, \vec{a}_{2}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{k}, \vec{b}_{1}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{k}, \vec{b}_{2}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{k}$ Note that the dr's of given lines are not proportional so, they are not parallel lines. The lines will be skew if they do not intersect each other also. Here $\displaystyle \vec{a}_{2}-\vec{a}_{1}=\hat{\mathrm{j}}-4 \hat{k}, \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{array}\right|=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{k}$ Consider $\displaystyle \left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)$ \[=(\hat{\mathrm{j}}-4 \hat{k}) \cdot(2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{k})=8 \neq 0 \] Hence lines will not intersect. So the lines are skew. Shortest Distance $\displaystyle =\frac{\left|\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)\right|}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}$ \[=\frac{8}{\sqrt{4+16+9}}=\frac{8}{\sqrt{29}} \] Let the wicket keeper divides the line segment in ratio $\displaystyle k: 1$ \[\begin{aligned} & \therefore \vec{\mathrm{W}}=\frac{k \vec{\mathrm{F}}+1 . \vec{\mathrm{B}}}{k+1} \\ & \Rightarrow 6 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}=\left(\frac{12 k+2}{k+1}\right) \hat{\mathrm{i}}+\left(\frac{18 k+8}{k+1}\right) \hat{\mathrm{j}} \\ & \Rightarrow k=\frac{2}{3} \end{aligned} \] $\displaystyle \mathrm{B}(2,8,0) \quad \mathrm{W}(6,12,0) \quad \mathrm{F}(12,18,0)$ Hence, the required ratio is $\displaystyle 2: 3$