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

To find the shortest distance between two skew lines given in vector form, we use the formula:

\(D = \frac{|(\mathbf{b_1} - \mathbf{b_2}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|}{|\mathbf{a_1} \times \mathbf{a_2}|}\)

Here:

• \(\mathbf{a_1}\) and \(\mathbf{a_2}\) are the direction vectors of the lines.
• \(\mathbf{b_1}\) and \(\mathbf{b_2}\) are position vectors of points on the lines.

 

Step-by-step Solution:

1. Identify direction vectors:
   • From the equation \(\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}\), the direction vector \(\mathbf{a_1}\) is \(2\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}\).
   • From the equation \(\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}\), the direction vector \(\mathbf{a_2}\) is \(2\mathbf{i} + 1\mathbf{j} - 3\mathbf{k}\).
2. Identify the position vectors of points on each line:
   • A point on the first line can be \((3, -15, 9)\) giving position vector: \(\mathbf{b_1} = 3\mathbf{i} - 15\mathbf{j} + 9\mathbf{k}\).
   • A point on the second line can be \((-1, 1, 9)\) giving position vector: \(\mathbf{b_2} = -1\mathbf{i} + 1\mathbf{j} + 9\mathbf{k}\).
3. Compute \(\mathbf{a_1} \times \mathbf{a_2}\):

The cross product is given by:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix}\)

Calculating the determinant, we have:

• \(= \mathbf{i}((-7)(-3) - (5)(1)) - \mathbf{j}((2)(-3) - (5)(2)) + \mathbf{k}((2)(1) - (-7)(2))\)
• \(= \mathbf{i}(21 - 5) - \mathbf{j}(-6 - 10) + \mathbf{k}(2 + 14)\)
• \(= 16\mathbf{i} + 16\mathbf{j} + 16\mathbf{k}\)

1. Calculate \(|\mathbf{a_1} \times \mathbf{a_2}|\):

\(|\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{16^2 + 16^2 + 16^2} = \sqrt{768} = 16\sqrt{3}\)

1. Compute \(\mathbf{b_1} - \mathbf{b_2}\):

\(= (3 + 1)\mathbf{i} + (-15 - 1)\mathbf{j} + (9 - 9)\mathbf{k}\)

\(= 4\mathbf{i} - 16\mathbf{j}\)

1. Calculate \(|(\mathbf{b_1} - \mathbf{b_2}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|\):

\(= |(4\mathbf{i} - 16\mathbf{j}) \cdot (16\mathbf{i} + 16\mathbf{j} + 16\mathbf{k})|\)

\(= |4 \times 16 + (-16) \times 16| = |64 - 256| = |-192| = 192\)

1. Therefore, the shortest distance is:

\(D = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\)

Therefore, the shortest distance between the lines is \(4\sqrt{3}\).

Answer 2

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: \[ S.D. = \frac{340}{\sqrt{953}} = 4\sqrt{3}. \]