Question

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

Hint:

Before applying the skew lines formula, quickly check if the direction vectors \(\vec{b_1}\) and \(\vec{b_2}\) are proportional. If they are, the lines are parallel, and a different formula for distance is used. The numerator of the formula is the scalar triple product \([\vec{a_2} - \vec{a_1} \quad \vec{b_1} \quad \vec{b_2}]\).

Answer 1

Step 1: Understanding the Concept:
The problem asks for the shortest distance between two lines in 3D space given in vector form. The lines are not parallel, so they are skew lines.
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:
\[ d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \] Step 3: Detailed Explanation:
From the given equations, we identify the vectors:
For the first line: \[ \vec{a_1} = \hat{i}+2\hat{j}+3\hat{k} \] \[ \vec{b_1} = \hat{i}-3\hat{j}+2\hat{k} \] For the second line: \[ \vec{a_2} = 4\hat{i}+5\hat{j}+6\hat{k} \] \[ \vec{b_2} = 2\hat{i}+3\hat{j}+\hat{k} \] First, we find the vector \(\vec{a_2} - \vec{a_1}\): \[ \vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i}+3\hat{j}+3\hat{k} \] Next, we calculate the cross product \(\vec{b_1} \times \vec{b_2}\): \[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -3 & 2
2 & 3 & 1 \end{vmatrix} \] \[ = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(3) - (-3)(2)) \] \[ = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 + 6) \] \[ = -9\hat{i} + 3\hat{j} + 9\hat{k} \] Now, we find the magnitude of the cross product: \[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} \] \(\sqrt{171} = \sqrt{9 \times 19} = 3\sqrt{19}\).
Next, we calculate the dot product \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\): \[ (3\hat{i}+3\hat{j}+3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = (3)(-9) + (3)(3) + (3)(9) \] \[ = -27 + 9 + 27 = 9 \] Finally, we substitute these values into the shortest distance formula: \[ d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}} \] Step 4: Final Answer:
The shortest distance between the two lines is \(\frac{3}{\sqrt{19}}\) units.