The shortest distance between the skew lines | vedclass

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(C) The shortest distance $d$ between two skew lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by the formu

Vedclass · Accepted Answer

$\sqrt{35}$

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(C) The shortest distance $d$ between two skew lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by the formula: $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)}{|\vec{b}_1 	imes \vec{b}_2|} ight|$.Given lines are $\vec{r}=(3 \hat{i}+4 \hat{j}-2 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}-7 \hat{j}-2 \hat{k})+\mu(\hat{i}+3 \hat{j}+2 \hat{k})$.Here,$\vec{a}_1 = 3 \hat{i}+4 \hat{j}-2 \hat{k}$,$\vec{b}_1 = -\hat{i}+2 \hat{j}+\hat{k}$,$\vec{a}_2 = \hat{i}-7 \hat{j}-2 \hat{k}$,and $\vec{b}_2 = \hat{i}+3 \hat{j}+2 \hat{k}$.First,calculate $\vec{a}_2 - \vec{a}_1 = (1-3)\hat{i} + (-7-4)\hat{j} + (-2-(-2))\hat{k} = -2\hat{i} - 11\hat{j}$.Next,calculate the cross product $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3\hat{j} - 5\hat{k}$.The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1+9+25} = \sqrt{35}$.Now,the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (-2\hat{i} - 11\hat{j} + 0\hat{k}) \cdot (\hat{i} + 3\hat{j} - 5\hat{k}) = (-2)(1) + (-11)(3) + (0)(-5) = -2 - 33 = -35$.Finally,$d = \left| \frac{-35}{\sqrt{35}} ight| = \sqrt{35}$.

The shortest distance between the skew lines $\vec{r}=(3 \hat{i}+4 \hat{j}-2 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}-7 \hat{j}-2 \hat{k})+\mu(\hat{i}+3 \hat{j}+2 \hat{k})$ is

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