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# 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) $\sqrt{6}$

Last updated date: 14th Sep 2024
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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.

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{align} & \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{align}
Now finding the value of ${{\vec{b}}_{1}}\times {{\vec{b}}_{2}}$ .
We need to find the cross product of the above vectors.
${{\bar{b}}_{1}}\times {{\bar{b}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 1 & 4 \\ 3 & 4 & 5 \\ \end{matrix} \right|$
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{align} & {{{\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{align}
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{align} & \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{121+4+25} \\ & =\sqrt{150} \end{align}
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{align} & \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{align}
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 $\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)$ .