Question

The shortest distance between the lines \[ \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{-1} \] and \[ \frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3} \] is:

• A. \( \frac{18}{\sqrt{5}} \)
• B. \( \frac{22}{3\sqrt{5}} \)
• C. \( \frac{46}{3\sqrt{5}} \)
• D. \( 6\sqrt{3} \)

Hint:

For finding the shortest distance between skew lines: - Use the determinant formula: \[ D = \frac{| (\vec{B} - \vec{A}) \cdot (\vec{d_1} \times \vec{d_2}) |}{|\vec{d_1} \times \vec{d_2}|}. \] - Ensure correct evaluation of vector cross product and dot product.

Answer 1

The Correct Option is A

Step 1: Identify direction vectors. 
The given lines are in symmetric form: \[ \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{-1} \] \[ \frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3} \] 
Direction vectors for the lines:
\[ \vec{d_1} = (2,3,-1), \quad \vec{d_2} = (2,1,3). \] A point on the first line is \( A(3,2,1) \) and a point on the second line is \( B(-3,6,5) \).
 Step 2: Compute the shortest distance formula. 
The shortest distance between two skew lines is given by: 
\[ D = \frac{| (\vec{B} - \vec{A}) \cdot (\vec{d_1} \times \vec{d_2}) |}{|\vec{d_1} \times \vec{d_2}|}. \] First, find \( \vec{B} - \vec{A} \): \[ \vec{B} - \vec{A} = (-3 - 3, 6 - 2, 5 - 1) = (-6,4,4). \] 
Step 3: Compute the cross product \( \vec{d_1} \times \vec{d_2} \). \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix} \] Expanding: \[ \hat{i} \begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} \] \[ = \hat{i} (3 \times 3 - (-1 \times 1)) - \hat{j} (2 \times 3 - (-1 \times 2)) + \hat{k} (2 \times 1 - 3 \times 2) \] \[ = \hat{i} (9 + 1) - \hat{j} (6 + 2) + \hat{k} (2 - 6) \] \[ = 10\hat{i} - 8\hat{j} - 4\hat{k}. \]
 Step 4: Compute the determinant \( (\vec{B} - \vec{A}) \cdot (\vec{d_1} \times \vec{d_2}) \). \[ (-6,4,4) \cdot (10,-8,-4) \] \[ = (-6 \times 10) + (4 \times -8) + (4 \times -4). \] \[ = -60 - 32 - 16 = -108. \] Taking the absolute value: \[ | -108 | = 108. \] 
Step 5: Compute \( |\vec{d_1} \times \vec{d_2}| \). \[ \sqrt{10^2 + (-8)^2 + (-4)^2} = \sqrt{100 + 64 + 16} = \sqrt{180} = 6\sqrt{5}. \] 
Step 6: Compute the shortest distance. \[ D = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}. \] Thus, the correct answer is: \[ \frac{18}{\sqrt{5}}. \]