Find the shortest distance between the lines | vedclass

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Question

(A) The given lines are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$.Rewriting

Vedclass · Accepted Answer

$\frac{8}{\sqrt{29}}$ units

Vedclass · Answer

(A) The given lines are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$.Rewriting the equations in the form $\vec{r} = \vec{a} + \lambda \vec{b}$:Line $1$: $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k})$Here,$\vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}$.Line $2$: $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})$Here,$\vec{a_2} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}$.The shortest distance $d$ between two skew lines is given by $d = \left| \frac{(\vec{b_1} 	imes \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} 	imes \vec{b_2}|} ight|$.First,calculate $\vec{b_1} 	imes \vec{b_2}$:$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & -2 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$.Magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.Next,calculate $\vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k}$.Now,$(\vec{b_1} 	imes \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = (2\hat{i} - 4\hat{j} - 3\hat{k}) \cdot (0\hat{i} + \hat{j} - 4\hat{k}) = (2)(0) + (-4)(1) + (-3)(-4) = 0 - 4 + 12 = 8$.Thus,$d = \left| \frac{8}{\sqrt{29}} ight| = \frac{8}{\sqrt{29}}$ units.

Find the shortest distance between the lines whose vector equations are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$.

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