Question

Find shortest distance between the sides of parrallelogram $\overline{r}=2\overline{i}-\overline{j}+\lambda (2\overline{i}+\overline{j}-3\overline{k})$ and $\overline{r}=\overline{i}-\overline{j}+2\overline{k}+\mu (2\overline{i}+\overline{j}-5\overline{k})$

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{5}}$
• C. $\frac{1}{\sqrt{7}}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{\sqrt{5}}$

the question has typing mistake both sides of parallelogram is parallel so

$\overrightarrow{r}=2\hat{i}-\hat{j}+\lambda (2\hat{i}+\hat{j}-5\hat{k})$

in place of

$\overrightarrow{r}=2\hat{i}-\hat{j}+\lambda (2\hat{i}+\hat{j}-3\hat{k})$

given sides

$\overrightarrow{r}=2\hat{i}-\hat{j}+\lambda (2\hat{i}+\hat{j}-5\hat{k})$

$\overrightarrow{r}=\hat{i}-\hat{j}+2\hat{k}+\mu (2\hat{i}+\hat{j}-5\hat{k})$

normal vector

$\overrightarrow{b}=2\hat{i}+\hat{j}-5\hat{k}$

posiotion vectors

$\overrightarrow{a_1}=2\hat{i}-\hat{j}$

$\overrightarrow{a_2}=\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-\hat{j}+2\hat{k}-2\hat{i}+\hat{j}$

$\overrightarrow{a_2}-\overrightarrow{a_1}=-\hat{i}+2\hat{k}$

$SD=\left|\right(\overrightarrow{a_2}-\overrightarrow{a_1})\times \hat{b}|$

$SD=\frac{\left|\right(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

$SD=\frac{\left|\right(-\hat{i}+2\hat{k})\times (2\hat{i}+\hat{j}-5\hat{k}\left)\right|}{\sqrt{2^2+1^2+(-5{)}^2}}$

$SD=\frac{\left|\right(-\hat{i}-3\hat{j}+2\hat{k})\times (2\hat{i}+\hat{j}-5\hat{k}\left)\right|}{\sqrt{4+1+25}}$

$SD=\frac{\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 0& 2\\ 2& 1& -5\end{array}\right|\right|}{\sqrt{30}}$

$SD=\frac{\left|\hat{i}\right(0-2)-\hat{j}(5-4)+\hat{k}(-1-0\left)\right|}{\sqrt{30}}$

$SD=\frac{|-2\hat{i}-\hat{j}-\hat{k}|}{\sqrt{30}}$

$SD=\frac{\sqrt{(-2{)}^2+(-1{)}^2+(-1{)}^2}}{\sqrt{30}}$

$SD=\frac{\sqrt{4+1+1}}{\sqrt{30}}$

$SD=\frac{\sqrt{6}}{\sqrt{30}}$

$SD=\frac{1}{\sqrt{5}}$