Question

The shortest distance between the line:
\[ \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5} \] and \[ \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1} \] is:

• A. \(\frac{187}{\sqrt{563}}\)
• B. \(\frac{178}{\sqrt{563}}\)
• C. \(\frac{185}{\sqrt{563}}\)
• D. \(\frac{179}{\sqrt{563}}\)

Answer 1

The Correct Option is A

The shortest distance between two skew lines is determined by a specific formula. Consider the following lines: \(L₁: \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5}\) and \(L₂: \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1}\).

The direction vectors for these lines are:

For \(L₁\): \(\mathbf{d₁} = (4, -11, 5)\)

For \(L₂\): \(\mathbf{d₂} = (3, -6, 1)\)

The formula for the distance \(d\) between two skew lines is: \(d = \frac{|(\mathbf{a₂} - \mathbf{a₁}) \cdot (\mathbf{d₁} \times \mathbf{d₂})|}{|\mathbf{d₁} \times \mathbf{d₂}|}\). The initial points on each line are \(\mathbf{a₁} = (3, -7, 1)\) and \(\mathbf{a₂} = (5, 9, -2)\).

The vector between points on the lines is calculated as: \(\mathbf{a₂} - \mathbf{a₁} = (5 - 3, 9 + 7, -2 - 1) = (2, 16, -3)\).

The cross product of \(\mathbf{d₁}\) and \(\mathbf{d₂}\) is: \(\mathbf{d₁} \times \mathbf{d₂} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix} = \mathbf{i}((-11)(1) - (5)(-6)) - \mathbf{j}((4)(1) - (5)(3)) + \mathbf{k}((4)(-6) - (-11)(3)) = \mathbf{i}(-11 + 30) - \mathbf{j}(4 - 15) + \mathbf{k}(-24 + 33) = (19, 11, 9)\).

The dot product \((\mathbf{a₂} - \mathbf{a₁}) \cdot (\mathbf{d₁} \times \mathbf{d₂})\) is: \((2, 16, -3) \cdot (19, 11, 9) = (2)(19) + (16)(11) + (-3)(9) = 38 + 176 - 27 = 187\).

The magnitude of the cross product \(|\mathbf{d₁} \times \mathbf{d₂}|\) is: \(\sqrt{19^2 + 11^2 + 9^2} = \sqrt{361 + 121 + 81} = \sqrt{563}\).

Finally, the shortest distance \(d\) is: \(d = \frac{|187|}{\sqrt{563}} = \frac{187}{\sqrt{563}}\).

The shortest distance between the given lines is \(\frac{187}{\sqrt{563}}\).