Find the shortest distance between the lines | vedclass

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(C) The given lines are $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\m

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$14$

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(C) The given lines are $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})$.First line: $\vec{r}=(8 \hat{i}-9 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k})$.So,$\vec{a}_{1}=8 \hat{i}-9 \hat{j}+10 \hat{k}$ and $\vec{b}_{1}=3 \hat{i}-16 \hat{j}+7 \hat{k}$.Second line: $\vec{a}_{2}=15 \hat{i}+29 \hat{j}+5 \hat{k}$ and $\vec{b}_{2}=3 \hat{i}+8 \hat{j}-5 \hat{k}$.The shortest distance $d$ is given by $d = \left|\frac{(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}_{1} 	imes \vec{b}_{2}|}ight|$.Calculate $\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -16 & 7 \ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80-56) - \hat{j}(-15-21) + \hat{k}(24+48) = 24 \hat{i}+36 \hat{j}+72 \hat{k}$.Magnitude $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{24^2 + 36^2 + 72^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84$.Calculate $\vec{a}_{2}-\vec{a}_{1} = (15-8)\hat{i} + (29-(-9))\hat{j} + (5-10)\hat{k} = 7\hat{i} + 38\hat{j} - 5\hat{k}$.Dot product $(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1}) = (24)(7) + (36)(38) + (72)(-5) = 168 + 1368 - 360 = 1176$.Shortest distance $d = \left|\frac{1176}{84}ight| = 14 	ext{ units}$.

Find the shortest distance between the lines given by $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})$. (in $\text{ units}$)

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