Question

The line $ L_1 $ is parallel to the vector $ \mathbf{a} = -3\hat{i} + 2\hat{j} + 4\hat{k} $ and passes through the point $ (7, 6, 2) $, and the line $ L_2 $ is parallel to the vector $ \mathbf{b} = 2\hat{i} + \hat{j} + 3\hat{k} $ and passes through the point $ (5, 3, 4) $. The shortest distance between the lines $ L_1 $ and $ L_2 $ is:

• A. \( \frac{23}{\sqrt{38}} \)
• B. \( \frac{21}{\sqrt{57}} \)
• C. \( \frac{23}{\sqrt{57}} \)
• D. \( \frac{21}{\sqrt{38}} \)

Hint:

For calculating the shortest distance between two skew lines, use the formula involving the cross product of direction vectors and the vector joining points on the lines. Be sure to calculate each component carefully.

Answer 1

The Correct Option is A

The parametric equations of the lines \( L_1 \) and \( L_2 \) are given as follows: For line \( L_1 \), passing through \( (7, 6, 2) \) and parallel to the vector \( \mathbf{a} = -3\hat{i} + 2\hat{j} + 4\hat{k} \): \[ L_1: (7 + \lambda(-3), 6 + \lambda(2), 2 + \lambda(4)) \quad \text{(Equation of line 1)} \] For line \( L_2 \), passing through \( (5, 3, 4) \) and parallel to the vector \( \mathbf{b} = 2\hat{i} + \hat{j} + 3\hat{k} \): \[ L_2: (5 + \lambda(3), 3 + \lambda(1), 4 + \lambda(3)) \quad \text{(Equation of line 2)} \] The shortest distance between skew lines is given by the formula: \[ d = \frac{| (\mathbf{b}_1 - \mathbf{b}_2) \cdot (\mathbf{a}_1 \times \mathbf{a}_2) |}{|\mathbf{a}_1 \times \mathbf{a}_2|} \] Where \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) are the direction vectors of the two lines, and \( \mathbf{b}_1 \) and \( \mathbf{b}_2 \) are points on the respective lines. Here, \( \mathbf{a}_1 = (-3, 2, 4) \), \( \mathbf{a}_2 = (2, 1, 3) \), \( \mathbf{b}_1 = (7, 6, 2) \), and \( \mathbf{b}_2 = (5, 3, 4) \). We compute the cross product \( \mathbf{a}_1 \times \mathbf{a}_2 \) first: \[ \mathbf{a}_1 \times \mathbf{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -3 & 2 & 4 2 & 1 & 3 \end{vmatrix} = (2 \times 3 - 4 \times 1) \hat{i} - (-3 \times 3 - 4 \times 2) \hat{j} + (-3 \times 1 - 2 \times 2) \hat{k} \] \[ = (6 - 4)\hat{i} - (-9 - 8)\hat{j} + (-3 - 4)\hat{k} \] \[ = 2 \hat{i} + 17 \hat{j} - 7 \hat{k} \] Now, the distance is calculated using the formula above: \[ d = \frac{|(2 \hat{i} + 17 \hat{j} - 7 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k})|}{\sqrt{342}} = \frac{69}{\sqrt{342}} = \frac{23}{\sqrt{38}} \] Thus, the shortest distance between the two lines is \( \frac{23}{\sqrt{38}} \).