Question

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

Hint:

The first step in any shortest distance problem between two lines is to check if the direction vectors are parallel. If they are, you must use the parallel line formula. If not, you use the formula for skew lines, which is \( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \). Using the wrong formula is a common mistake.

Answer 1

Step 1: Understanding the Concept:
The given equations are of the form \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \). We first check if the lines are parallel by comparing their direction vectors, \( \vec{b}_1 \) and \( \vec{b}_2 \). If they are parallel (\(\vec{b}_1 = \vec{b}_2 = \vec{b}\)), we use the formula for the shortest distance between parallel lines.
Step 2: Key Formula or Approach:
The direction vectors are \( \vec{b}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k} \) and \( \vec{b}_2 = 2\hat{i} + 3\hat{j} + 6\hat{k} \). Since \( \vec{b}_1 = \vec{b}_2 \), the lines are parallel.
The formula for the shortest distance between two parallel lines is: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \] where \( \vec{a}_1 \) and \( \vec{a}_2 \) are position vectors of points on the lines and \( \vec{b} \) is the common direction vector.
Step 3: Detailed Explanation:
From the given equations, we have: \[ \vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k} \] \[ \vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} \] \[ \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \] First, calculate the vector difference \( \vec{a}_2 - \vec{a}_1 \): \[ \vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} \] \[ \vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k} \] Next, calculate the cross product \( (\vec{a}_2 - \vec{a}_1) \times \vec{b} \): \[ (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & -1
2 & 3 & 6 \end{vmatrix} \] \[ = \hat{i}((1)(6) - (-1)(3)) - \hat{j}((2)(6) - (-1)(2)) + \hat{k}((2)(3) - (1)(2)) \] \[ = \hat{i}(6 + 3) - \hat{j}(12 + 2) + \hat{k}(6 - 2) \] \[ = 9\hat{i} - 14\hat{j} + 4\hat{k} \] Now, find the magnitude of this cross product: \[ |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81 + 196 + 16} = \sqrt{293} \] Then, find the magnitude of the direction vector \( \vec{b} \): \[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] Finally, calculate the shortest distance: \[ d = \frac{\sqrt{293}}{7} \] Step 4: Final Answer:
The shortest distance between the two parallel lines is \( \frac{\sqrt{293}}{7} \) units.