If p is the shortest distance between the lin | vedclass

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(C) Step $1$: Identify points and vectors for both lines.Line $1$: $\vec{a_1} = -\hat{i} - \hat{j} - \hat{k}$, $\vec{b_1} = 7\hat{i} - 6\hat{j} + \hat

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$10$

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(C) Step $1$: Identify points and vectors for both lines.Line $1$: $\vec{a_1} = -\hat{i} - \hat{j} - \hat{k}$, $\vec{b_1} = 7\hat{i} - 6\hat{j} + \hat{k}$.Line $2$: $\vec{a_2} = 3\hat{i} + 5\hat{j} + 7\hat{k}$, $\vec{b_2} = \hat{i} - 2\hat{j} + \hat{k}$.Step $2$: Calculate $\vec{a_2} - \vec{a_1} = (3 - (-1))\hat{i} + (5 - (-1))\hat{j} + (7 - (-1))\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k}$.Step $3$: Calculate $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & -6 & 1 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-6 + 2) - \hat{j}(7 - 1) + \hat{k}(-14 + 6) = -4\hat{i} - 6\hat{j} - 8\hat{k}$.Step $4$: Calculate magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$.Step $5$: Shortest distance $p = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|} = \frac{|(4\hat{i} + 6\hat{j} + 8\hat{k}) \cdot (-4\hat{i} - 6\hat{j} - 8\hat{k})|}{2\sqrt{29}} = \frac{|-16 - 36 - 64|}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$.Step $6$: Since $\sqrt{29} \approx 5.385$, $p = 2 	imes 5.385 = 10.77$. Thus, $[p] = [10.77] = 10$.

If $p$ is the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$, then $[p]$ is... (where $[.]$ denotes the greatest integer function.)

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