Question

Find the shortest distance between the lines
\(\vec{r} = (4 + \lambda)\hat{i + (2\lambda - 1)\hat{j} - 3\lambda\hat{k}\)
\(\vec{r} = (1 + 2\mu)\hat{i} + (4\mu - 1)\hat{j} + (2 - 5\mu)\hat{k}\)}

Hint:

If \(\vec{b_1} \times \vec{b_2} = \vec{0}\), the lines are parallel and a different formula is used.
In Line 2, be careful to collect terms for \(\hat{i}, \hat{j}, \text{and } \hat{k}\) correctly from the combined form.

Answer 1

Concept:
The shortest distance between two skew lines \[ \vec{r}=\vec{a_1}+\lambda\vec{b_1} \] and \[ \vec{r}=\vec{a_2}+\mu\vec{b_2} \] is given by: \[ d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} \] 
Step 1: Identify the vectors
From the first line: \[ \vec{r}=(4\hat{i}-\hat{j})+\lambda(\hat{i}+2\hat{j}-3\hat{k}) \] Therefore: \[ \vec{a_1}=4\hat{i}-\hat{j}, \qquad \vec{b_1}=\hat{i}+2\hat{j}-3\hat{k} \] From the second line: \[ \vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}+4\hat{j}-5\hat{k}) \] Therefore: \[ \vec{a_2}=\hat{i}-\hat{j}+2\hat{k}, \qquad \vec{b_2}=2\hat{i}+4\hat{j}-5\hat{k} \] 
Step 2: Calculate the required vectors
First: \[ \vec{a_2}-\vec{a_1} =(1-4)\hat{i}+(-1+1)\hat{j}+(2-0)\hat{k} \] \[ \vec{a_2}-\vec{a_1}=-3\hat{i}+2\hat{k} \] Now calculate the cross product: \[ \vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} \] \[ =\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4) \] \[ \vec{b_1}\times\vec{b_2}=2\hat{i}-\hat{j} \] 
Step 3: Apply the shortest distance formula
The numerator is: \[ |(-3\hat{i}+2\hat{k})\cdot(2\hat{i}-\hat{j})| \] \[ =|(-3)(2)+(0)(-1)+(2)(0)| \] \[ =|-6|=6 \] The denominator is: \[ |\vec{b_1}\times\vec{b_2}| =\sqrt{2^2+(-1)^2} \] \[ =\sqrt{5} \] Therefore: \[ d=\frac{6}{\sqrt{5}} \] 
Final Answer:
The shortest distance between the two lines is: \[ \boxed{\frac{6}{\sqrt{5}}\text{ units}} \]