Question

The shortest distance between the lines
\[\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}\]and
\[\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}\] is:

• A. \( 6\sqrt{3} \)
• B. \( 4\sqrt{3} \)
• C. \( 5\sqrt{3} \)
• D. \( 8\sqrt{3} \)

Answer 1

The Correct Option is B

The shortest distance between the given lines is calculated as:

\[ S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}. \]

1. Step 1: Extract points and direction vectors: From the first line: \[ \vec{a}_1 = (3, -15, 9), \quad \vec{b}_1 = (2, -7, 5). \] From the second line: \[ \vec{a}_2 = (-1, 1, 9), \quad \vec{b}_2 = (2, 1, -3). \]
2. Step 2: Compute \(\vec{b}_1 \times \vec{b}_2\): \[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix}. \] Expanding the determinant: \[ \vec{b}_1 \times \vec{b}_2 = \hat{i}[(16) - (-5)] - \hat{j}[(10) - (-6)] + \hat{k}[(2) - (-14)]. \] Simplify: \[ \vec{b}_1 \times \vec{b}_2 = 21\hat{i} - 16\hat{j} + 16\hat{k}. \]
3. Step 3: Magnitude of \(\vec{b}_1 \times \vec{b}_2\): \[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{21^2 + (-16)^2 + 16^2}. \] Simplify: \[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{441 + 256 + 256} = \sqrt{953}. \]
4. Step 4: Find \(\vec{a}_2 - \vec{a}_1\): \[ \vec{a}_2 - \vec{a}_1 = (-1 - 3, 1 - (-15), 9 - 9) = (-4, 16, 0). \]
5. Step 5: Dot product \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\): \[ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4)(21) + (16)(-16) + (0)(16). \] Simplify: \[ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = -84 - 256 + 0 = -340. \]
6. Step 6: Shortest distance: Substitute into the formula: \[ S.D. = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|-340|}{\sqrt{953}}. \] Simplify: \[ S.D. = \frac{340}{\sqrt{953}} = 4\sqrt{3}. \]