Question

Define skew lines. Using only the vector approach, find the shortest distance between the following two skew lines.
$\begin{array}{rl}& \overrightarrow{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}\\ & \overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})\end{array}$

Answer 1

Hint: The shortest distance between two skew line equations ${\overrightarrow{r}}_1={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and ${\overrightarrow{r}}_2={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ is $\left|{\displaystyle \frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2).({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}}\right|$. The vector $({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)$ can be found by $({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ b_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\end{array}\right|$ , where $b_{11},b_{12},b_{13}$ are the components of ${\overrightarrow{b}}_1$ and $b_{21},b_{22},b_{23}$ are the components of ${\overrightarrow{b}}_2$ respectively.
Substituting all the values we can get the shortest distance.

Complete step-by-step answer:
We can define skew lines as follows,
Skew lines are straight lines in a three-dimensional form which are not parallel and do not cross.
Let’s consider the first skew line to be ${\overrightarrow{r}}_1=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}....................(i)$
Let’s consider the second skew line to be $${\overrightarrow{r}}_2=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})....................(ii)$$
As we can see that both the equations are in different forms. Let us convert equation (i) in equation (ii) we get,
${\overrightarrow{r}}_1=8\hat{i}-9\hat{j}+10\hat{k}+\lambda (3\hat{i}-16\hat{j}+7\hat{k})....................(iii)$
Now, comparing equation (ii) and equation (iii) with ${\overrightarrow{r}}_1={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and ${\overrightarrow{r}}_2={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ we get the values to be as follows,
$\begin{array}{rl}& {\overrightarrow{a}}_1=8\hat{i}-9\hat{j}+10\hat{k}...................(iv)\\ & {\overrightarrow{b}}_1=3\hat{i}-16\hat{j}+7\hat{k}...................(v)\\ & {\overrightarrow{a}}_2=15\hat{i}+29\hat{j}+5\hat{k}...................(vi)\\ & {\overrightarrow{b}}_2=3\hat{i}+8\hat{j}-5\hat{k}...................(vii)\end{array}$
The shortest distance between two skew line equations ${\overrightarrow{r}}_1={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and ${\overrightarrow{r}}_2={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ is $\left|{\displaystyle \frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2).({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}}\right|$
Let us first find the value of $({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)$. Substituting the values from equation (vi) and (iv) we get,
$({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(15\hat{i}+29\hat{j}+5\hat{k})-(8\hat{i}-9\hat{j}+10\hat{k})$
Solving the equation, we get,
$\begin{array}{rl}& ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(15-8)\hat{i}+(29-(-9))\hat{j}+(5-10)\hat{k}\\ & ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=7\hat{i}+38\hat{j}-5\hat{k}.....................(viii)\end{array}$
Now, let's find the value of $({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)$ . Substituting the values from the equation (v) and (vii) we get,
$({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|$
Solving the above equation we get,
$\begin{array}{rl}& ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)=[(-16\times -5)-(7\times 8)]\hat{i}-[(3\times -5)-(7\times 3)]\hat{j}+[(3\times 8)-(3\times -16)]\hat{k}\\ & ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)=[80-56]\hat{i}-[-15-21]\hat{j}+[24+48]\hat{k}\\ & ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)=24\hat{i}+36\hat{j}+72\hat{k}....................\left(ix\right)\end{array}$
Now, let's find the value of $\left|({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\right|$.
The magnitude of the $({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)$ can be found as follows,
$\left|({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\right|=\sqrt{{\left(\text{Coefficient of }\hat{i}\right)}^2+{\left(\text{Coefficient of }\hat{j}\right)}^2+{\left(\text{Coefficient of }\hat{k}\right)}^2}$
 Finding the magnitude of the vector we get,
$\left|({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\right|=|24\hat{i}+36\hat{j}+72\hat{k}|$
Solving the above equation we get,
$\begin{array}{rl}& \left|({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\right|=\sqrt{{24}^2+{36}^2+{72}^2}\\ & \left|({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\right|=\sqrt{7056}\\ & \left|({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\right|=84.....................(x)\end{array}$
Combining all the values and substituting in the formula we get,
The shortest distance $=\left|{\displaystyle \frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2).({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}}\right|=\left|{\displaystyle \frac{(24\hat{i}+36\hat{j}+48\hat{k}).(7\hat{i}+38\hat{j}-5\hat{k})}{84}}\right|$
In the numerator, we need to find the dot product of two vectors. The formula for finding the dot product is as follows,
$$\left(\overrightarrow{a}\right).\left(\overrightarrow{b}\right)=(a_x\times b_x)+(a_y\times b_y)+(a_z\times b_z)$$ where, $a_x,a_y,a_z$ are the x, y, z components of $\overrightarrow{a}$ and $b_x,b_y,b_z$ are the x, y, z components of $\overrightarrow{b}$ .
Solving the equation further,
$$=\left|{\displaystyle \frac{(24\times 7)+(36\times 38)+(48\times -5)}{84}}\right|=\left|{\displaystyle \frac{192+1368-240}{84}}\right|={\displaystyle \frac{110}{7}}\text{units}$$ .
Therefore, the distance between the given lines is ${\displaystyle \frac{110}{7}}$units.


Note: It is easily confused while interpreting the components of the ${\overrightarrow{a}}_1,{\overrightarrow{a}}_2,{\overrightarrow{b}}_1,{\overrightarrow{b}}_2$ . If the final answer is negative, we need to write only the magnitude of the answer obtained. Another common mistake which can be made is that there is a by default negative sign while finding the $j^{th}$ component of $({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)$ . This is one of the rules while finding the determinant which needs to be taken care of.