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 for lines \( L_1 \) and \( L_2 \) are provided.Line \( L_1 \) passes through \( (7, 6, 2) \) and is parallel to \( \mathbf{a} = -3\hat{i} + 2\hat{j} + 4\hat{k} \). Its equation is:\[L_1: (7 - 3\lambda, 6 + 2\lambda, 2 + 4\lambda)\]Line \( L_2 \) passes through \( (5, 3, 4) \) and is parallel to \( \mathbf{b} = 2\hat{i} + \hat{j} + 3\hat{k} \). Its equation is:\[L_2: (5 + 3\lambda, 3 + \lambda, 4 + 3\lambda)\]The formula for the shortest distance between skew lines is:\[d = \frac{| (\mathbf{b}_1 - \mathbf{b}_2) \cdot (\mathbf{a}_1 \times \mathbf{a}_2) |}{|\mathbf{a}_1 \times \mathbf{a}_2|}\]In this formula, \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) are the direction vectors of the lines, and \( \mathbf{b}_1 \) and \( \mathbf{b}_2 \) are points on the lines.We have \( \mathbf{a}_1 = (-3, 2, 4) \), \( \mathbf{a}_2 = (2, 1, 3) \), \( \mathbf{b}_1 = (7, 6, 2) \), and \( \mathbf{b}_2 = (5, 3, 4) \).First, we calculate the cross product \( \mathbf{a}_1 \times \mathbf{a}_2 \):\[\mathbf{a}_1 \times \mathbf{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -3 & 2 & 4 2 & 1 & 3 \end{vmatrix}= (2 \times 3 - 4 \times 1) \hat{i} - (-3 \times 3 - 4 \times 2) \hat{j} + (-3 \times 1 - 2 \times 2) \hat{k}\]\[= (6 - 4)\hat{i} - (-9 - 8)\hat{j} + (-3 - 4)\hat{k}\]\[= 2 \hat{i} + 17 \hat{j} - 7 \hat{k}\]Next, we compute the distance using the formula:\[d = \frac{|(7, 6, 2) - (5, 3, 4) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k})|}{|2 \hat{i} + 17 \hat{j} - 7 \hat{k}|}\]Which simplifies to:\[d = \frac{|(2, 3, -2) \cdot (2, 17, -7)|}{\sqrt{2^2 + 17^2 + (-7)^2}} = \frac{|(2 \times 2) + (3 \times 17) + (-2 \times -7)|}{\sqrt{4 + 289 + 49}}\]\[d = \frac{|4 + 51 + 14|}{\sqrt{342}} = \frac{70}{\sqrt{342}} = \frac{35}{\sqrt{171}} = \frac{35}{3\sqrt{19}}\]There appears to be a miscalculation in the original text. Let's recompute the dot product part of the numerator:\[(\mathbf{b}_1 - \mathbf{b}_2) = (7-5, 6-3, 2-4) = (2, 3, -2)\]\[(\mathbf{b}_1 - \mathbf{b}_2) \cdot (\mathbf{a}_1 \times \mathbf{a}_2) = (2, 3, -2) \cdot (2, 17, -7) = (2)(2) + (3)(17) + (-2)(-7) = 4 + 51 + 14 = 69\]So the distance is:\[d = \frac{|69|}{\sqrt{342}} = \frac{69}{\sqrt{342}}\]Simplifying the fraction:\[\frac{69}{\sqrt{342}} = \frac{69}{\sqrt{9 \times 38}} = \frac{69}{3\sqrt{38}} = \frac{23}{\sqrt{38}}\]The shortest distance between the two lines is \( \frac{23}{\sqrt{38}} \).