If r=b+ta and r=d+sc are two skew lines, then | vedclass

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(D) Given the two skew lines are $r = b + ta$ and $r = d + sc$. The formula for the shortest distance between two skew lines $r = a_1 + \lambda b_1$ a

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Orthogonal projection of $(b - d)$ on $(a 	imes c)$.

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(D) Given the two skew lines are $r = b + ta$ and $r = d + sc$. The formula for the shortest distance between two skew lines $r = a_1 + \lambda b_1$ and $r = a_2 + \mu b_2$ is given by $d = \left| \frac{(a_2 - a_1) \cdot (b_1 	imes b_2)}{|b_1 	imes b_2|} ight|$. Here, $a_1 = b$, $b_1 = a$, $a_2 = d$, and $b_2 = c$. Substituting these values, the shortest distance is $\left| \frac{(d - b) \cdot (a 	imes c)}{|a 	imes c|} ight|$. Since $|x| = |-x|$, this is equivalent to $\left| \frac{(b - d) \cdot (a 	imes c)}{|a 	imes c|} ight|$. This expression represents the magnitude of the orthogonal projection of the vector $(b - d)$ onto the vector $(a 	imes c)$.

If $r=b+ta$ and $r=d+sc$ are two skew lines, then the shortest distance between them is

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