Question

Find the shortest distance between lines $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda \left( 2\hat{i}+\hat{j}+4\hat{k} \right)$ and $\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu \left( 3\hat{i}+4\hat{j}+5\hat{k} \right)$ .
(a) 2
(b) $\dfrac{1}{6}$
(c) $\dfrac{1}{\sqrt{6}}$
(d) $\sqrt{6}$

Answer 1

Hint : In order to solve this problem, we need to know the formula for the shortest distance. The formula for shortest distance of lines represented by $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}}$ is as follows,
Shortest distance = $\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$ .We can calculate the denominator by cross-product and solve the numerator by the dot product.

Complete step-by-step answer :
We need to find the shortest distance between two lines.
Let's compare the first equation $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda \left( 2\hat{i}+\hat{j}+4\hat{k} \right)$ to $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$
We get,
${{\vec{a}}_{1}}=\hat{i}+2\hat{j}+3\hat{k}$
${{\bar{b}}_{1}}=2\hat{i}+\hat{j}+4\hat{k}$
Now comparing the second equation $\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu \left( 3\hat{i}+4\hat{j}+5\hat{k} \right)$ with $\vec{r}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}}$ we get,
${{\bar{a}}_{2}}=2\hat{i}+4\hat{j}+5\hat{k}$
${{\bar{b}}_{2}}=3\hat{i}+4\hat{j}+5\hat{k}$
The formula for shortest distance of lines represented by $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}}$ is as follows,
Shortest distance = $\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|..............................(i)$
Now we need to find the value of $\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)$ ,
Substituting the values, we get,
$\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)=\left( 2\hat{i}+4\hat{j}+5\hat{k} \right)-\left( \hat{i}+2\hat{j}+3\hat{k} \right)$
Solving we get,
$\begin{aligned} & \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)=\left( 2-1 \right)\hat{i}+\left( 4-2 \right)\hat{j}+\left( 5-3 \right)\hat{k} \\\ & =\hat{i}+2\hat{j}+2\hat{k}..................................(ii) \end{aligned}$
Now finding the value of ${{\vec{b}}_{1}}\times {{\vec{b}}_{2}}$ .
We need to find the cross product of the above vectors.

{\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 2 & 1 & 4 \\\ 3 & 4 & 5 \\\ \end{matrix} \right|$$ Where along the second row we have the components of $ {{\vec{b}}_{1}} $ and along the third row we have the components of $ {{\vec{b}}_{2}} $ . Solving this we get, $ \begin{aligned} & {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left( 5-16 \right)\hat{i}-\left( 10-12 \right)\hat{j}+\left( 8-3 \right)\hat{k} \\\ & =-11\hat{i}+2\hat{j}+5\hat{k}.................................(iii) \end{aligned} $ We also need to find the magnitude of $ {{\vec{b}}_{1}}\times {{\vec{b}}_{2}} $ that is $ \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right| $ . Therefore, we get, $ \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{{{\left( -11 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 5 \right)}^{2}}} $ This is obtained by squaring all the terms, adding them up and taking the square root. Solving this we get, $ \begin{aligned} & \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{121+4+25} \\\ & =\sqrt{150} \end{aligned} $ Substituting all the values in equation (i), we get, Shortest distance = $ \left| \dfrac{\left( -11\hat{i}+2\hat{j}+5\hat{k} \right).\left( \hat{i}+2\hat{j}+2\hat{k} \right)}{\sqrt{150}} \right| $ Solving this and taking the dot product in the numerator we get, Shortest distance = $ \left| \dfrac{\left( -11\hat{i}+2\hat{j}+5\hat{k} \right).\left( \hat{i}+2\hat{j}+2\hat{k} \right)}{\sqrt{150}} \right| $ In dot product, we need to add the multiplication of the $ {{i}^{th}} $ , $ {{j}^{th}} $ and the $ {{k}^{th}} $ component. Solving this we get, $ \begin{aligned} & \text{Shortest distance}=\left| \dfrac{-11+4+10}{\sqrt{150}} \right| \\\ & =\dfrac{3}{\sqrt{150}} \\\ & =\dfrac{\sqrt{3}\times \sqrt{3}}{\sqrt{3\times 5\times 2\times 5}} \\\ & =\dfrac{\sqrt{3}}{\sqrt{5\times 2\times 5}} \\\ & =\dfrac{\sqrt{3}}{5\sqrt{2}} \\\ & =\dfrac{\sqrt{3}\times \sqrt{2}}{5\sqrt{2}\times \sqrt{2}} \\\ & =\dfrac{\sqrt{6}}{5\times 2} \\\ & =\dfrac{\sqrt{6}}{10} \end{aligned} $ Hence the shortest distance between two lines is $ \dfrac{\sqrt{6}}{10} $ units. **Note** : We need to be careful while performing the cross product there is a negative sign in the $ \hat{j} $ part. We have asked to find the distance therefore, we need to take the modulus of all the scalar terms. Also, while calculating $ \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right) $ and not $ \left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right) $ .