Question

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

Answer 1

Hint: The shortest distance between two skew line equations ${{\vec{r}}_{1}}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and ${{\vec{r}}_{2}}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}}$ is $\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$. The vector $\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)$ can be found by $\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{b}_{11}} & {{b}_{12}} & {{b}_{13}} \\\ {{b}_{21}} & {{b}_{22}} & {{b}_{23}} \\\ \end{matrix} \right|$ , where ${{b}_{11}},{{b}_{12}},{{b}_{13}}$ are the components of ${{\vec{b}}_{1}}$ and ${{b}_{21}},{{b}_{22}},{{b}_{23}}$ are the components of ${{\vec{b}}_{2}}$ 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 ${{\vec{r}}_{1}}=\left( 8+3\lambda \right)\hat{i}-\left( 9+16\lambda \right)\hat{j}+\left( 10+7\lambda \right)\hat{k}....................(i)$
Let’s consider the second skew line to be ${{\vec{r}}_{2}}=15\hat{i}+29\hat{j}+5\hat{k}+\mu \left( 3\hat{i}+8\hat{j}-5\hat{k} \right)....................(ii)$
As we can see that both the equations are in different forms. Let us convert equation (i) in equation (ii) we get,
${{\vec{r}}_{1}}=8\hat{i}-9\hat{j}+10\hat{k}+\lambda \left( 3\hat{i}-16\hat{j}+7\hat{k} \right)....................(iii)$
Now, comparing equation (ii) and equation (iii) with ${{\vec{r}}_{1}}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and ${{\vec{r}}_{2}}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}}$ we get the values to be as follows,
$\begin{aligned} & {{{\vec{a}}}_{1}}=8\hat{i}-9\hat{j}+10\hat{k}...................(iv) \\\ & {{{\vec{b}}}_{1}}=3\hat{i}-16\hat{j}+7\hat{k}...................(v) \\\ & {{{\vec{a}}}_{2}}=15\hat{i}+29\hat{j}+5\hat{k}...................(vi) \\\ & {{{\vec{b}}}_{2}}=3\hat{i}+8\hat{j}-5\hat{k}...................(vii) \\\ \end{aligned}$
The shortest distance between two skew line equations ${{\vec{r}}_{1}}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and ${{\vec{r}}_{2}}={{\vec{a}}_{2}}+\mu {{\vec{b}}_{2}}$ is $\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$
Let us first find the value of $\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)$. Substituting the values from equation (vi) and (iv) we get,
$\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)=\left( 15\hat{i}+29\hat{j}+5\hat{k} \right)-\left( 8\hat{i}-9\hat{j}+10\hat{k} \right)$
Solving the equation, we get,
$\begin{aligned} & \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)=\left( 15-8 \right)\hat{i}+\left( 29-\left( -9 \right) \right)\hat{j}+\left( 5-10 \right)\hat{k} \\\ & \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)=7\hat{i}+38\hat{j}-5\hat{k}.....................(viii) \\\ \end{aligned}$
Now, let's find the value of $\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)$ . Substituting the values from the equation (v) and (vii) we get,
$\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 3 & -16 & 7 \\\ 3 & 8 & -5 \\\ \end{matrix} \right|$
Solving the above equation we get,
$\begin{aligned} & \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)=\left[ \left( -16\times -5 \right)-\left( 7\times 8 \right) \right]\hat{i}-\left[ \left( 3\times -5 \right)-\left( 7\times 3 \right) \right]\hat{j}+\left[ \left( 3\times 8 \right)-\left( 3\times -16 \right) \right]\hat{k} \\\ & \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)=\left[ 80-56 \right]\hat{i}-\left[ -15-21 \right]\hat{j}+\left[ 24+48 \right]\hat{k} \\\ & \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)=24\hat{i}+36\hat{j}+72\hat{k}....................\left( ix \right) \\\ \end{aligned}$
Now, let's find the value of $\left| \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right) \right|$.
The magnitude of the $\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)$ can be found as follows,
$\left| \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right) \right|=\sqrt{{{\left( \text{Coefficient of }\hat{i} \right)}^{2}}+{{\left( \text{Coefficient of }\hat{j} \right)}^{2}}+{{\left( \text{Coefficient of }\hat{k} \right)}^{2}}}$
Finding the magnitude of the vector we get,
$\left| \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right) \right|=\left| 24\hat{i}+36\hat{j}+72\hat{k} \right|$
Solving the above equation we get,
$\begin{aligned} & \left| \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right) \right|=\sqrt{{{24}^{2}}+{{36}^{2}}+{{72}^{2}}} \\\ & \left| \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right) \right|=\sqrt{7056} \\\ & \left| \left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right) \right|=84.....................(x) \\\ \end{aligned}$
Combining all the values and substituting in the formula we get,
The shortest distance $=\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|=\left| \dfrac{\left( 24\hat{i}+36\hat{j}+48\hat{k} \right).\left( 7\hat{i}+38\hat{j}-5\hat{k} \right)}{84} \right|$
In the numerator, we need to find the dot product of two vectors. The formula for finding the dot product is as follows,
$\left( {\vec{a}} \right).\left( {\vec{b}} \right)=\left( {{a}_{x}}\times {{b}_{x}} \right)+\left( {{a}_{y}}\times {{b}_{y}} \right)+\left( {{a}_{z}}\times {{b}_{z}} \right)$ where, ${{a}_{x}},{{a}_{y}},{{a}_{z}}$ are the x, y, z components of $\vec{a}$ and ${{b}_{x}},{{b}_{y}},{{b}_{z}}$ are the x, y, z components of $\vec{b}$ .
Solving the equation further,
$=\left| \dfrac{\left( 24\times 7 \right)+\left( 36\times 38 \right)+\left( 48\times -5 \right)}{84} \right|=\left| \dfrac{192+1368-240}{84} \right|=\dfrac{110}{7}\text{units}$ .
Therefore, the distance between the given lines is $\dfrac{110}{7}$units.

Note: It is easily confused while interpreting the components of the ${{\vec{a}}_{1}},{{\vec{a}}_{2}},{{\vec{b}}_{1}},{{\vec{b}}_{2}}$ . 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 $\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)$ . This is one of the rules while finding the determinant which needs to be taken care of.