Find the shortest distance between lines r=(2 i-j)+ (2 i+j-3 k) and r=i-j+2 k+ (2 i+j-5 k)

The shortest distance between lines r=a_1+ _1 b_1 and r=a_2+ _2 b_2 is | (a_2-a_1 ) (b_1 b_2 ) |b_1 b_2 | | Here a_1=2 i-j, a_2=i-j+2 k, b_1=2 i+j+3 k, b_2=2 i+j-5 k a_2-a_1=(i-j+2 k)-(2 i-j)=-i+2 k And b_1 b_2= | ccc i & j & k 2 & 1 & -3 2 & 1 & -5 |=-2 i+4 j & |b_1 b_2 |=4+16=20=2 5 & (a_2-a_1 ) (b_1 b_2 )=(-i+2 k) (-2 i+4 j)=2 The required shortest distance | (a_2-a_1 ) (b_1 b_2 ) |b_1 b_2 | |= |2 2 5 |=1 5 unit.

Find the shortest distance between lines r=(2 i-j)+ (2 i+j-3 k) a…

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Question

Find the shortest distance between lines $\bar{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=\hat{i}-\hat{j}+2 \hat{k}+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$

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Answer

The shortest distance between lines $\bar{r}=\bar{a}_1+\lambda_1 \bar{b}_1$ and $\bar{r}=\bar{a}_2+\lambda_2 \bar{b}_2$ is $\left|\frac{\left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)}{\left|\bar{b}_1 \times \bar{b}_2\right|}\right|$
Here $\bar{a}_1=2 \hat{i}-\hat{j}, \bar{a}_2=\hat{i}-\hat{j}+2 \hat{k}, \bar{b}_1=2 \hat{i}+\hat{j}+3 \hat{k}, \bar{b}_2=2 \hat{i}+\hat{j}-5 \hat{k}$
$
\overline{a_2}-\overline{a_1}=(\hat{i}-\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j})=-\hat{i}+2 \hat{k}
$
And $\bar{b}_1 \times \bar{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & 1 & -5\end{array}\right|=-2 \hat{i}+4 \hat{j}$
$
\begin{aligned}
\therefore \quad & \left|\bar{b}_1 \times \bar{b}_2\right|=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5} \\
& \left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=(-\hat{i}+2 \hat{k}) \cdot(-2 \hat{i}+4 \hat{j})=2
\end{aligned}
$
The required shortest distance $\left|\frac{\left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)}{\left|\bar{b}_1 \times \bar{b}_2\right|}\right|=\left|\frac{2}{2 \sqrt{5}}\right|=\frac{1}{\sqrt{5}}$ unit.