Find the shortest distance between the lines | vedclass

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(A) The given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.The shortest distance $d$ between two skew

Vedclass · Accepted Answer

$\frac{3}{\sqrt{19}}$ units

Vedclass · Answer

(A) The given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.The shortest distance $d$ between two skew lines is given by the formula:$d = \left| \frac{(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1})}{|\vec{b}_{1} 	imes \vec{b}_{2}|} ight|$Comparing the given equations,we have:$\vec{a}_{1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b}_{1} = \hat{i} - 3\hat{j} + 2\hat{k}$$\vec{a}_{2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$,$\vec{b}_{2} = 2\hat{i} + 3\hat{j} + \hat{k}$Calculate $\vec{a}_{2} - \vec{a}_{1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.Calculate the cross product $\vec{b}_{1} 	imes \vec{b}_{2}$:$\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 2 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(1-4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k}$.Calculate the magnitude $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$.Calculate the dot product $(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1}) = (-9\hat{i} + 3\hat{j} + 9\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = -27 + 9 + 27 = 9$.Thus,$d = \left| \frac{9}{3\sqrt{19}} ight| = \frac{3}{\sqrt{19}}$ units.

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

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