The shortest distance between the skew lines | vedclass

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(A) The shortest distance between two skew lines $r=a+tb$ and $r=c+sd$ is given by the formula: $	ext{Shortest distance} = \left| \frac{(c-a) \cdot (

Vedclass · Accepted Answer

$9$

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(A) The shortest distance between two skew lines $r=a+tb$ and $r=c+sd$ is given by the formula: $	ext{Shortest distance} = \left| \frac{(c-a) \cdot (b 	imes d)}{|b 	imes d|} ight|$.Given lines are $r=(6 \hat{i}+2 \hat{j}+2 \hat{k})+t(\hat{i}-2 \hat{j}+2 \hat{k})$ and $r=(-4 \hat{i}-\hat{k})+s(3 \hat{i}-2 \hat{j}-2 \hat{k})$.Here,$a=6 \hat{i}+2 \hat{j}+2 \hat{k}$,$b=\hat{i}-2 \hat{j}+2 \hat{k}$,$c=-4 \hat{i}-\hat{k}$,and $d=3 \hat{i}-2 \hat{j}-2 \hat{k}$.First,calculate $c-a = (-4 \hat{i}-\hat{k}) - (6 \hat{i}+2 \hat{j}+2 \hat{k}) = -10 \hat{i}-2 \hat{j}-3 \hat{k}$.Next,calculate the cross product $b 	imes d = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) = 8 \hat{i}+8 \hat{j}+4 \hat{k}$.The magnitude $|b 	imes d| = \sqrt{8^2+8^2+4^2} = \sqrt{64+64+16} = \sqrt{144} = 12$.The dot product $(c-a) \cdot (b 	imes d) = (-10 \hat{i}-2 \hat{j}-3 \hat{k}) \cdot (8 \hat{i}+8 \hat{j}+4 \hat{k}) = -80 - 16 - 12 = -108$.Therefore,the shortest distance is $\left| \frac{-108}{12} ight| = |-9| = 9$.

The shortest distance between the skew lines $r=(6 \hat{i}+2 \hat{j}+2 \hat{k})+t(\hat{i}-2 \hat{j}+2 \hat{k})$ and $r=(-4 \hat{i}-\hat{k})+s(3 \hat{i}-2 \hat{j}-2 \hat{k})$ is

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