Find the shortest distance between lines vec{ | vedclass

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(A) The given lines are:$\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$ ...........$(1)$$\vec{r} = -4\hat{i} - \ha

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$9$

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(A) The given lines are:$\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$ ...........$(1)$$\vec{r} = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$ ...........$(2)$The shortest distance $d$ between two lines $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$ is given by:$d = \left| \frac{(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1})}{|\vec{b}_{1} 	imes \vec{b}_{2}|} ight|$ ...........$(3)$Comparing the given equations with the standard form,we get:$\vec{a}_{1} = 6\hat{i} + 2\hat{j} + 2\hat{k}$,$\vec{b}_{1} = \hat{i} - 2\hat{j} + 2\hat{k}$$\vec{a}_{2} = -4\hat{i} - \hat{k}$,$\vec{b}_{2} = 3\hat{i} - 2\hat{j} - 2\hat{k}$Now,calculate $\vec{a}_{2} - \vec{a}_{1} = (-4\hat{i} - \hat{k}) - (6\hat{i} + 2\hat{j} + 2\hat{k}) = -10\hat{i} - 2\hat{j} - 3\hat{k}$.Calculate the cross product $\vec{b}_{1} 	imes \vec{b}_{2}$:$\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6)) = 8\hat{i} + 8\hat{j} + 4\hat{k}$.The magnitude $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.Now,calculate the dot product $(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1})$:$(8\hat{i} + 8\hat{j} + 4\hat{k}) \cdot (-10\hat{i} - 2\hat{j} - 3\hat{k}) = (8)(-10) + (8)(-2) + (4)(-3) = -80 - 16 - 12 = -108$.Finally,the shortest distance $d = \left| \frac{-108}{12} ight| = |-9| = 9$ units.

Find the shortest distance between lines $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and $\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$. (in $units$)

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