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 \( \frac{23}{\sqrt{38}} \).

Answer 2

This problem asks for the shortest distance between two skew lines, \( L_1 \) and \( L_2 \), in three-dimensional space. We are given a point on each line and a vector parallel to each line.

Concept Used:

The shortest distance between two skew lines, \( L_1 \) passing through point \( \mathbf{p}_1 \) with direction vector \( \mathbf{d}_1 \), and \( L_2 \) passing through point \( \mathbf{p}_2 \) with direction vector \( \mathbf{d}_2 \), is given by the formula:

\[ d = \frac{| (\mathbf{p}_2 - \mathbf{p}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) |}{| \mathbf{d}_1 \times \mathbf{d}_2 |} \]

This formula calculates the projection of the vector connecting the two points onto the vector that is perpendicular to both lines (the cross product of their direction vectors).

Step-by-Step Solution:

Step 1: Identify the points and direction vectors for each line from the problem statement.

For line \( L_1 \):

• It passes through the point \( P_1 = (7, 6, 2) \). The position vector is \( \mathbf{p}_1 = 7\hat{i} + 6\hat{j} + 2\hat{k} \).
• It is parallel to the vector \( \mathbf{a} \). The direction vector is \( \mathbf{d}_1 = -3\hat{i} + 2\hat{j} + 4\hat{k} \).

For line \( L_2 \):

• It passes through the point \( P_2 = (5, 3, 4) \). The position vector is \( \mathbf{p}_2 = 5\hat{i} + 3\hat{j} + 4\hat{k} \).
• It is parallel to the vector \( \mathbf{b} \). The direction vector is \( \mathbf{d}_2 = 2\hat{i} + \hat{j} + 3\hat{k} \).

Step 2: Calculate the vector connecting the points \( P_1 \) and \( P_2 \), which is \( \mathbf{p}_2 - \mathbf{p}_1 \).

\[ \mathbf{p}_2 - \mathbf{p}_1 = (5\hat{i} + 3\hat{j} + 4\hat{k}) - (7\hat{i} + 6\hat{j} + 2\hat{k}) \] \[ \mathbf{p}_2 - \mathbf{p}_1 = (5-7)\hat{i} + (3-6)\hat{j} + (4-2)\hat{k} \] \[ \mathbf{p}_2 - \mathbf{p}_1 = -2\hat{i} - 3\hat{j} + 2\hat{k} \]

Step 3: Calculate the cross product of the direction vectors, \( \mathbf{d}_1 \times \mathbf{d}_2 \).

\[ \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} \] \[ = \hat{i}(2 \cdot 3 - 4 \cdot 1) - \hat{j}((-3) \cdot 3 - 4 \cdot 2) + \hat{k}((-3) \cdot 1 - 2 \cdot 2) \] \[ = \hat{i}(6 - 4) - \hat{j}(-9 - 8) + \hat{k}(-3 - 4) \] \[ = 2\hat{i} - (-17)\hat{j} - 7\hat{k} \] \[ \mathbf{d}_1 \times \mathbf{d}_2 = 2\hat{i} + 17\hat{j} - 7\hat{k} \]

Step 4: Calculate the scalar triple product, \( (\mathbf{p}_2 - \mathbf{p}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) \).

\[ (-2\hat{i} - 3\hat{j} + 2\hat{k}) \cdot (2\hat{i} + 17\hat{j} - 7\hat{k}) \] \[ = (-2)(2) + (-3)(17) + (2)(-7) \] \[ = -4 - 51 - 14 = -69 \]

The absolute value is \( |-69| = 69 \).

Step 5: Calculate the magnitude of the cross product, \( |\mathbf{d}_1 \times \mathbf{d}_2| \).

\[ |\mathbf{d}_1 \times \mathbf{d}_2| = |2\hat{i} + 17\hat{j} - 7\hat{k}| \] \[ = \sqrt{(2)^2 + (17)^2 + (-7)^2} \] \[ = \sqrt{4 + 289 + 49} = \sqrt{342} \]

Step 6: Substitute the values from Step 4 and Step 5 into the shortest distance formula.

\[ d = \frac{| (\mathbf{p}_2 - \mathbf{p}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) |}{| \mathbf{d}_1 \times \mathbf{d}_2 |} \] \[ d = \frac{69}{\sqrt{342}} \]

To simplify the radical, we can factor 342: \( 342 = 9 \times 38 \). Therefore, \( \sqrt{342} = \sqrt{9 \times 38} = 3\sqrt{38} \).

\[ d = \frac{69}{3\sqrt{38}} = \frac{23}{\sqrt{38}} \]

The shortest distance between the lines \( L_1 \) and \( L_2 \) is \( \frac{23}{\sqrt{38}} \).