The line L_1 is parallel to the vector vec{a} | vedclass

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(A) The equations of the lines are $L_1: \vec{r} = (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k})$ and $L_2: \vec{r

Vedclass · Accepted Answer

$\frac{23}{\sqrt{38}}$

Vedclass · Answer

(A) The equations of the lines are $L_1: \vec{r} = (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k})$ and $L_2: \vec{r} = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) + \mu(2 \hat{i} + \hat{j} + 3 \hat{k})$.Let $\vec{a_1} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k}$,$\vec{a_2} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$,$\vec{v_1} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$,and $\vec{v_2} = 2 \hat{i} + \hat{j} + 3 \hat{k}$.The shortest distance $d$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} 	imes \vec{v_2})|}{|\vec{v_1} 	imes \vec{v_2}|}$.First,calculate $\vec{a_2} - \vec{a_1} = (5-7)\hat{i} + (3-6)\hat{j} + (4-2)\hat{k} = -2 \hat{i} - 3 \hat{j} + 2 \hat{k}$.Next,calculate $\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 2 & 4 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-4) - \hat{j}(-9-8) + \hat{k}(-3-4) = 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$.The magnitude $|\vec{v_1} 	imes \vec{v_2}| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{4 + 289 + 49} = \sqrt{342} = \sqrt{9 	imes 38} = 3 \sqrt{38}$.The dot product $|(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} 	imes \vec{v_2})| = |(-2)(2) + (-3)(17) + (2)(-7)| = |-4 - 51 - 14| = |-69| = 69$.Thus,$d = \frac{69}{3 \sqrt{38}} = \frac{23}{\sqrt{38}}$.

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