Question

The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is

• A. $2 \sqrt{29}$
• B. 1
• C. $\sqrt{\frac{37}{29}}$
• D. $\frac{\sqrt{29}}{2}$

Answer 1

The Correct Option is A

To find the shortest distance between two skew lines, we can use the formula for the shortest distance between two lines in space. Given the lines:

Line 1: \(\frac{x+7}{-6}=\frac{y-6}{7}=z\)

Line 2: \(\frac{7-x}{2}=y-2=z-6\)

Both lines are in the symmetric form. We can rewrite them in the parametric form:

• For Line 1, let \(t=z\):
  • \(x = -6t - 7\)
  • \(y = 7t + 6\)
  • \(z = t\)
• For Line 2, let \(s=z-6\):
  • \(x = 7 - 2s\)
  • \(y = s + 2\)
  • \(z = s + 6\)

The direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\) of the lines are:

• \(\mathbf{d}_1 = \langle -6, 7, 1 \rangle\) for Line 1
• \(\mathbf{d}_2 = \langle -2, 1, 1 \rangle\) for Line 2

Find a point on each line:

• For Line 1 (choose \(t = 0\)): \((x_1, y_1, z_1) = (-7, 6, 0)\)
• For Line 2 (choose \(s = 0\)): \((x_2, y_2, z_2) = (7, 2, 6)\)

The shortest distance \((d)\) between the skew lines is given by:

\(d = \frac{|(\mathbf{r}_2 - \mathbf{r}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|}\)

where \(\mathbf{r}_1 = \langle -7, 6, 0 \rangle\) and \(\mathbf{r}_2 = \langle 7, 2, 6 \rangle\).

Calculate \(\mathbf{r}_2 - \mathbf{r}_1\):

\(\mathbf{r}_2 - \mathbf{r}_1 = \langle 7 - (-7), 2 - 6, 6 - 0 \rangle = \langle 14, -4, 6 \rangle\)

Calculate \(\mathbf{d}_1 \times \mathbf{d}_2\):

\(\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6 & 7 & 1 \\ -2 & 1 & 1 \end{vmatrix}\)

\(= \mathbf{i}(7\cdot1 - 1\cdot1) - \mathbf{j}(-6\cdot1 - 1\cdot(-2)) + \mathbf{k}(-6\cdot1 - 7\cdot(-2))\)

\(= \langle 6, 10, 20 \rangle\)

Compute the magnitude:

\(|\mathbf{d}_1 \times \mathbf{d}_2| = \sqrt{6^2 + 10^2 + 20^2} = \sqrt{36 + 100 + 400} = \sqrt{536} = 2 \times \sqrt{134}\)

Compute the scalar triple product and its magnitude:

\((\mathbf{r}_2 - \mathbf{r}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) = \langle 14, -4, 6 \rangle \cdot \langle 6, 10, 20 \rangle\)

\(= 14 \times 6 + (-4) \times 10 + 6 \times 20 = 84 - 40 + 120 = 164\)

\(|164| = 164\)

Shortest Distance:

\(d = \frac{164}{2 \times \sqrt{134}}\)

After simplification: \(d = 2 \sqrt{29}\)

Hence, the shortest distance between the lines is \(2 \sqrt{29}\).