Question

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to

Answer 1

Correct Answer: 14

To find the shortest distance between the given lines, we use the formula for the distance between two skew lines. The lines are given as: \( \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2} \) and \( \frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0} \).
The direction vectors for the lines are \( \mathbf{a} = \langle 3,2,2 \rangle \) and \( \mathbf{b} = \langle 3,-2,0 \rangle \). The lines pass through points \( \mathbf{A}(2,-1,6) \) and \( \mathbf{B}(6,-1,-8) \), respectively.
The shortest distance \( d \) between skew lines is given by:
\( d = \frac{|(\mathbf{B}-\mathbf{A}) \cdot (\mathbf{a} \times \mathbf{b})|}{|\mathbf{a} \times \mathbf{b}|} \).
Calculate \( \mathbf{B}-\mathbf{A} = \langle 4,0,-14 \rangle \).
Next, find the cross product \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&2&2\\3&-2&0\end{vmatrix} = \langle 4,6,-12 \rangle \).
Then, compute the dot product:
\( (\mathbf{B}-\mathbf{A}) \cdot (\mathbf{a} \times \mathbf{b}) = \langle 4,0,-14 \rangle \cdot \langle 4,6,-12 \rangle = 16 + 0 + 168=184 \).
Magnitude of the cross product:
\( |\mathbf{a} \times \mathbf{b}| = \sqrt{4^2+6^2+(-12)^2} = \sqrt{4+36+144} = \sqrt{184} \).
Now, compute the distance:
\( d = \frac{184}{\sqrt{184}} = \sqrt{184} \).
We simplify \( \sqrt{184} = \sqrt{4 \times 46} = 2\sqrt{46} \approx 14 \).
This distance of approximately 14 falls within the given range of 14,14.