Q74 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

The shortest distance between the skew-lines $\mathbf{r}=(-\hat{\mathbf{i}}+3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$ and $\mathbf{r}=(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is

  • A
    $\frac{10}{\sqrt{17}}$
  • Correct answer
    B
    $\frac{22}{\sqrt{17}}$
  • C
    9
  • D
    8

Solution

Given, lines $\mathbf{r}=(-\hat{\mathbf{i}}+3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$ and $\mathbf{r}=(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ Shortest distance between lines is $\begin{aligned} & \mathrm{SD}=\frac{((3+1) \hat{\mathbf{i}}+\hat{\mathbf{j}}-(\hat{\mathbf{k}}+3) \hat{\mathbf{k}})-[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})}{\times(2 \mathbf{i}-\mathbf{j}+2 \mathbf{k})]} \\ & \mathrm{SD}=\frac{(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{i}}) \cdot(12 \hat{\mathbf{j}}+6 \hat{\mathbf{i}})|\times|(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}) \mid}{|12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}|} \\ & \mathrm{SD}=\frac{48+8+32}{\sqrt{144+64+64}}=\frac{88}{4 \sqrt{17}}=\frac{22}{\sqrt{17}} \end{aligned}$

Hence, the correct answer is option: 'B'.

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