Question

Find the shortest distance between the following pairs of lines whose vector equations are: $\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k})$ and

$\vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})$

Accepted Answer

$\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})$ Comparing the

given equations with the equations

$\vec{r} = \vec{a_{1}} + λ \vec{b_{1}}$ and $\vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$

$\vec{a_{1}} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k}$

$\vec{a_{2}} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k}$

$\vec{b_{1}} = 3 \hat{i} - \hat{j} + \hat{k}$

$\vec{b_{2}} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$

$∴ \vec{a_{2}} - \vec{a_{1}} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}$

and $\vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix}|$

$= - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}$

$\Rightarrow |\vec{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 6)}^{2} + {(- 15)}^{2} + 3^{2}}$

$= \sqrt{36 + 225 + 9}$

$= \sqrt{270}$

$(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (- 6 \hat{i} - 15 \hat{j} + 3 \hat{k}) \cdot (- 6 \hat{i} - 15 \hat{j} + 3 \hat{k})$

$= 36 + 225 + 9$

$= 270$

The shortest distance between the lines

$\vec{r} = \vec{a_{1}} + λ \vec{b_{1}}$ and $\vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by

$d = |\frac{(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}})}{|\vec{b_{1}} \times \vec{b_{2}}|}|$

$= \frac{270}{\sqrt{270}}$

$= \sqrt{270}$

R D Sharma Solutions for Chapter: Straight Line in Space, Exercise 5: EXERCISE

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Straight Line in Space, Exercise 5: EXERCISE

Questions from R D Sharma Solutions for Chapter: Straight Line in Space, Exercise 5: EXERCISE with Hints & Solutions

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