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If $\mathbf{r}=\mathbf{b}+t \mathbf{a}, \mathbf{r}=\mathbf{d}+s \mathbf{c}$ are the two skew lines, then the shortest distance between them is
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Verified Answer
The correct answer is:
Orthogonal projection of (b - d) on (a x c).
Given, $\mathbf{r}=\mathbf{b}+$ ta and $\mathbf{r}=\mathbf{d}+s \mathbf{c}$
$\begin{aligned}
\therefore \quad \text { S. D. } & =\left|\frac{(\mathbf{b}-\mathbf{d}) \cdot(\mathbf{a} \times \mathbf{c})}{|\mathbf{a} \times \mathbf{c}|}\right| \\
& =\text { projection of }(\mathbf{b}-\mathbf{d}) \text { on }(\mathbf{a} \times \mathbf{c}) .
\end{aligned}$
$\begin{aligned}
\therefore \quad \text { S. D. } & =\left|\frac{(\mathbf{b}-\mathbf{d}) \cdot(\mathbf{a} \times \mathbf{c})}{|\mathbf{a} \times \mathbf{c}|}\right| \\
& =\text { projection of }(\mathbf{b}-\mathbf{d}) \text { on }(\mathbf{a} \times \mathbf{c}) .
\end{aligned}$
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