The shortest distance between the skew lines $\mathbf{r}=(6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2…

1. 9
2. $\frac{40}{7}$
3. 108
4. 120

Solution

Given, $\begin{array}{ll} & \mathbf{r}=(6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ \text { and } & \mathbf{r}=(-4 \hat{\mathbf{i}}-\hat{\mathbf{k}})+s(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ \text { Here, } & \mathbf{a}=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathbf{c}=-4 \hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{d}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\end{array}$
$ \begin{aligned} & \text { Now, } \quad \mathbf{b} \times \mathbf{d}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right| \\ & =\hat{\mathbf{i}}(4+4)-\hat{\mathbf{j}}(-2-6)+\hat{\mathbf{k}}(-2+6)=8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned} $ By Eq. (i), $ \begin{aligned} & =\left|\frac{(10 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})}{\sqrt{64+64+16}}\right| \\ & =\left|\frac{80+16+12}{\sqrt{144}}\right|=\frac{108}{12}=9 \end{aligned} $

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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