Question

Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ and $\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$.

Hint:

Remember: Shortest distance (skew lines) = $\frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.

Answer 1

Concept: The shortest distance between two skew lines is given by: \[ \text{Distance} = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \] where $\vec{a_1}, \vec{a_2}$ are points on the lines and $\vec{b_1}, \vec{b_2}$ are direction vectors. Step 1: {Identify vectors.}
\[ \vec{a_1} = (1,2,3), \quad \vec{b_1} = (1,-3,2) \] \[ \vec{a_2} = (4,5,6), \quad \vec{b_2} = (2,3,1) \]
Step 2: {Compute } $\vec{a_2} - \vec{a_1}$.
\[ \vec{a_2} - \vec{a_1} = (3,3,3) \]
Step 3: {Find cross product } $\vec{b_1} \times \vec{b_2}$.
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \] \[ = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(3) - (-3)(2)) \] \[ = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 + 6) \] \[ = (-9)\hat{i} + 3\hat{j} + 9\hat{k} \]
Step 4: {Dot product.}
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (3,3,3) \cdot (-9,3,9) \] \[ = -27 + 9 + 27 = 9 \]
Step 5: {Magnitude of cross product.}
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} \]
Step 6: {Final distance.}
\[ \text{Distance} = \frac{|9|}{\sqrt{171}} = \frac{9}{\sqrt{171}} \]
Step 7: {Simplified form.}
\[ \text{Distance} = \frac{3}{\sqrt{19}} \]
Step 8: {Conclusion.}
Thus, the shortest distance between the given lines is $\frac{3}{\sqrt{19}}$.