Question

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

Hint:

The numerator of the distance formula, \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\), is a scalar triple product, which can also be computed as the determinant \(\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1
b_{1x} & b_{1y} & b_{1z}
b_{2x} & b_{2y} & b_{2z} \end{vmatrix}\). This can sometimes be a quicker calculation.

Answer 1

Step 1: Understanding the Concept:
The two given lines are in vector form \(\vec{r} = \vec{a} + t\vec{b}\). Since their direction vectors are not proportional, the lines are skew (or intersecting). The shortest distance between two skew lines is the length of the perpendicular line segment connecting them.
Step 2: Key Formula or Approach:
The shortest distance \(d\) 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 the formula: \[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \] Step 3: Detailed Explanation:
From the given equations of the lines, we identify the vectors: \[ \vec{a_1} = \hat{i} + \hat{j} + 0\hat{k} \quad \text{and} \quad \vec{b_1} = 2\hat{i} - \hat{j} + \hat{k} \] \[ \vec{a_2} = 2\hat{i} + \hat{j} - \hat{k} \quad \text{and} \quad \vec{b_2} = 3\hat{i} - 5\hat{j} + 2\hat{k} \] First, calculate the vector connecting the two points on the lines, \(\vec{a_2} - \vec{a_1}\): \[ \vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} + 0\hat{j} - \hat{k} \] Next, calculate the cross product of the direction vectors, \(\vec{b_1} \times \vec{b_2}\): \[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
3 & -5 & 2 \end{vmatrix} = \hat{i}(-2 - (-5)) - \hat{j}(4 - 3) + \hat{k}(-10 - (-3)) \] \[ \vec{b_1} \times \vec{b_2} = 3\hat{i} - \hat{j} - 7\hat{k} \] Now, find the magnitude of this cross product: \[ |\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \] Next, calculate the scalar triple product for the numerator of the formula: \[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} - \hat{k}) \cdot (3\hat{i} - \hat{j} - 7\hat{k}) \] \[ = (1)(3) + (0)(-1) + (-1)(-7) = 3 + 0 + 7 = 10 \] Finally, substitute these values into the shortest distance formula: \[ d = \frac{|10|}{\sqrt{59}} = \frac{10}{\sqrt{59}} \] Step 4: Final Answer:
The shortest distance between the two lines is \(\frac{10}{\sqrt{59}}\) units.