Question

Find the shortest distance between the lines: \[ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} \] and \[ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}. \]

Hint:

For skew lines, use the formula \( D = \frac{|(\mathbf{Q} - \mathbf{P}) \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|} \).

Answer 1

To determine the shortest distance between two skew lines given in symmetric form, the following steps are followed.

1. Line Parameter Extraction:

Line 1:
The symmetric equations are:

\[ \frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3} \]

Direction vector: \[ \vec{d_1} = \langle 2, 1, -3 \rangle \] Point on the line (taking parameter = 0): \[ A(-1, 1, 9) \]

Line 2:
The symmetric equations are:

\[ \frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5} \]

Direction vector: \[ \vec{d_2} = \langle 2, -7, 5 \rangle \] Point on the line (taking parameter = 0): \[ B(3, -15, 9) \]

2. Vector Between the Given Points:

\[ \vec{AB} = \vec{B} - \vec{A} = \langle 3 - (-1), -15 - 1, 9 - 9 \rangle = \langle 4, -16, 0 \rangle \]

3. Formula for Shortest Distance:

\[ \text{Distance} = \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|} {|\vec{d_1} \times \vec{d_2}|} \]

4. Cross Product Calculation:

\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} \]

\[ = \mathbf{i}(5 - 21) - \mathbf{j}(10 + 6) + \mathbf{k}(-14 - 2) = \langle -16, -16, -16 \rangle \]

5. Scalar Triple Product:

\[ \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 4(-16) + (-16)(-16) + 0(-16) = 192 \]

6. Magnitude of Cross Product:

\[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(-16)^2 + (-16)^2 + (-16)^2} = \sqrt{768} = 16\sqrt{3} \]

7. Final Distance:

\[ \text{Distance} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

Final Result:

\[ \boxed{4\sqrt{3}\ \text{units}} \]