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 solve the problem, we are given two skew lines in symmetric form. We are to find the shortest distance between them.

1. Identify Direction Vectors and Points on Each Line:

First line:
\[ \frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3} \] Direction vector: \( \mathbf{d_1} = \langle 2, 1, -3 \rangle \)
Point on the line: set the common ratio = 0 → \( x = -1, y = 1, z = 9 \)
So, point \( A = (-1, 1, 9) \)

Second line:
\[ \frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5} \] Direction vector: \( \mathbf{d_2} = \langle 2, -7, 5 \rangle \)
Point on the line: set the common ratio = 0 → \( x = 3, y = -15, z = 9 \)
So, point \( B = (3, -15, 9) \)

2. Vector Joining Points A and B:
\[ \vec{AB} = \vec{B} - \vec{A} = \langle 3 - (-1), -15 - 1, 9 - 9 \rangle = \langle 4, -16, 0 \rangle \]

3. Use the Formula for Shortest Distance Between Skew Lines:
\[ \text{Distance} = \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \]

4. Compute the Cross Product \( \vec{d_1} \times \vec{d_2} \):
\[ \vec{d_1} = \langle 2, 1, -3 \rangle,\quad \vec{d_2} = \langle 2, -7, 5 \rangle \] \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = \mathbf{i}(1 \cdot 5 - (-3) \cdot (-7)) - \mathbf{j}(2 \cdot 5 - (-3) \cdot 2) + \mathbf{k}(2 \cdot (-7) - 1 \cdot 2) \] \[ = \mathbf{i}(5 - 21) - \mathbf{j}(10 + 6) + \mathbf{k}(-14 - 2) = \langle -16, -16, -16 \rangle \]

5. Compute Dot Product \( \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) \):
\[ \vec{AB} = \langle 4, -16, 0 \rangle,\quad \vec{d_1} \times \vec{d_2} = \langle -16, -16, -16 \rangle \] \[ \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 4 \cdot (-16) + (-16) \cdot (-16) + 0 \cdot (-16) = -64 + 256 + 0 = 192 \]

6. Magnitude of the Cross Product:
\[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(-16)^2 + (-16)^2 + (-16)^2} = \sqrt{3 \cdot 256} = \sqrt{768} = 16\sqrt{3} \]

7. Final Calculation:
\[ \text{Distance} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

Final Answer:
The shortest distance between the lines is \( \boxed{4\sqrt{3}} \) units.