Question

Find the shortest distance between lines $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$ and $\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu (3\hat{i}+4\hat{j}+5\hat{k})$ .
(a) 2
(b) ${\displaystyle \frac{1}{6}}$
(c) ${\displaystyle \frac{1}{\sqrt{6}}}$
(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 $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ is as follows,
Shortest distance = $\left|{\displaystyle \frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2).({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}}\right|$ .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 $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$ to $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$
We get,
 ${\overrightarrow{a}}_1=\hat{i}+2\hat{j}+3\hat{k}$
$${\overline{b}}_1=2\hat{i}+\hat{j}+4\hat{k}$$
Now comparing the second equation $\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu (3\hat{i}+4\hat{j}+5\hat{k})$ with $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ we get,
$${\overline{a}}_2=2\hat{i}+4\hat{j}+5\hat{k}$$
$${\overline{b}}_2=3\hat{i}+4\hat{j}+5\hat{k}$$
The formula for shortest distance of lines represented by $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ is as follows,
Shortest distance = $\left|{\displaystyle \frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2).({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}}\right|..............................(i)$
Now we need to find the value of $({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)$ ,
 Substituting the values, we get,
 $({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(2\hat{i}+4\hat{j}+5\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$
Solving we get,
 $\begin{array}{rl}& ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(2-1)\hat{i}+(4-2)\hat{j}+(5-3)\hat{k}\\ & =\hat{i}+2\hat{j}+2\hat{k}..................................(ii)\end{array}$
Now finding the value of ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2$ .
We need to find the cross product of the above vectors.
$${\overline{b}}_1\times {\overline{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 4\\ 3& 4& 5\end{array}\right|$$
Where along the second row we have the components of ${\overrightarrow{b}}_1$ and along the third row we have the components of ${\overrightarrow{b}}_2$ .
Solving this we get,
 $\begin{array}{rl}& {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=(5-16)\hat{i}-(10-12)\hat{j}+(8-3)\hat{k}\\ & =-11\hat{i}+2\hat{j}+5\hat{k}.................................(iii)\end{array}$
We also need to find the magnitude of ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2$ that is $|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|$ .
Therefore, we get,
  $|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{{(-11)}^2+{\left(2\right)}^2+{\left(5\right)}^2}$
This is obtained by squaring all the terms, adding them up and taking the square root.
Solving this we get,
 $\begin{array}{rl}& |{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{121+4+25}\\ & =\sqrt{150}\end{array}$
Substituting all the values in equation (i), we get,
Shortest distance = $\left|{\displaystyle \frac{(-11\hat{i}+2\hat{j}+5\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{150}}}\right|$
Solving this and taking the dot product in the numerator we get,
Shortest distance = $\left|{\displaystyle \frac{(-11\hat{i}+2\hat{j}+5\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{150}}}\right|$
In dot product, we need to add the multiplication of the $i^{th}$ , $j^{th}$ and the $k^{th}$ component.
Solving this we get,
 $\begin{array}{rl}& \text{Shortest distance}=\left|{\displaystyle \frac{-11+4+10}{\sqrt{150}}}\right|\\ & ={\displaystyle \frac{3}{\sqrt{150}}}\\ & ={\displaystyle \frac{\sqrt{3}\times \sqrt{3}}{\sqrt{3\times 5\times 2\times 5}}}\\ & ={\displaystyle \frac{\sqrt{3}}{\sqrt{5\times 2\times 5}}}\\ & ={\displaystyle \frac{\sqrt{3}}{5\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{3}\times \sqrt{2}}{5\sqrt{2}\times \sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{6}}{5\times 2}}\\ & ={\displaystyle \frac{\sqrt{6}}{10}}\end{array}$
Hence the shortest distance between two lines is ${\displaystyle \frac{\sqrt{6}}{10}}$ 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 $({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)$ and not $({\overrightarrow{a}}_1-{\overrightarrow{a}}_2)$ .