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 formula for the shortest distance between two skew lines in vector form is:

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

Where:

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

 

Step-by-step Calculation:

1. Determine the direction vectors:
   • For the line \(\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}\), the direction vector \(\mathbf{a_1}\) is \(2\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}\).
   • For the line \(\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}\), 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 is \((3, -15, 9)\), yielding the position vector \(\mathbf{b_1} = 3\mathbf{i} - 15\mathbf{j} + 9\mathbf{k}\).
   • A point on the second line is \((-1, 1, 9)\), yielding the position vector \(\mathbf{b_2} = -1\mathbf{i} + 1\mathbf{j} + 9\mathbf{k}\).
3. Calculate the cross product \(\mathbf{a_1} \times \mathbf{a_2}\):

The cross product is computed as the determinant:

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

Evaluating the determinant yields:

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

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

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

1. Compute the difference of position vectors \(\mathbf{b_1} - \mathbf{b_2}\):

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

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

1. Calculate the absolute value of the dot product \(|(\mathbf{b_1} - \mathbf{b_2}) \cdot (\mathbf{a_1} \times \mathbf{a_2})|\):

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

\(= |(4)(16) + (-16)(16) + (0)(16)| = |64 - 256| = |-192| = 192\)

1. The shortest distance D is computed as:

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

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