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Question: Answered & Verified by Expert
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
MathematicsThree Dimensional GeometryTS EAMCETTS EAMCET 2020 (14 Sep Shift 2)
Options:
  • A $\frac{10}{\sqrt{17}}$
  • B $\frac{22}{\sqrt{17}}$
  • C 9
  • D 8
Solution:
2698 Upvotes Verified Answer
The correct answer is: $\frac{22}{\sqrt{17}}$
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
$\begin{aligned}
& \mathrm{SD}=\frac{((3+1) \hat{\mathbf{i}}+\hat{\mathbf{j}}-(\hat{\mathbf{k}}+3) \hat{\mathbf{k}})-[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})}{\times(2 \mathbf{i}-\mathbf{j}+2 \mathbf{k})]} \\
& \mathrm{SD}=\frac{(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{i}}) \cdot(12 \hat{\mathbf{j}}+6 \hat{\mathbf{i}})|\times|(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}) \mid}{|12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}|} \\
& \mathrm{SD}=\frac{48+8+32}{\sqrt{144+64+64}}=\frac{88}{4 \sqrt{17}}=\frac{22}{\sqrt{17}}
\end{aligned}$

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