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(A) The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $

Vedclass · Accepted Answer

$\sqrt{35}$

Vedclass · Answer

(A) The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$.Here,$\vec{a_1} = 3\hat{i} + 4\hat{j} - 2\hat{k}$ and $\vec{a_2} = \hat{i} - 7\hat{j} - 2\hat{k}$.$\vec{a_2} - \vec{a_1} = (1-3)\hat{i} + (-7-4)\hat{j} + (-2 - (-2))\hat{k} = -2\hat{i} - 11\hat{j} + 0\hat{k}$.$\vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3\hat{j} - 5\hat{k}$.$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (-2\hat{i} - 11\hat{j} + 0\hat{k}) \cdot (\hat{i} + 3\hat{j} - 5\hat{k}) = (-2)(1) + (-11)(3) + (0)(-5) = -2 - 33 = -35$.$d = \frac{|-35|}{\sqrt{35}} = \frac{35}{\sqrt{35}} = \sqrt{35}$.

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