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

Step 1: Identify the direction vectors. The given lines are in symmetric form: \[ \text{Line 1: } \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}. \] Direction vector of line 1: \[ \mathbf{d_1} = (2, 1, -3). \] Point on line 1: \( P(-1,1,9) \). \[ \text{Line 2: } \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}. \] Direction vector of line 2: \[ \mathbf{d_2} = (2, -7, 5). \] Point on line 2: \( Q(3,-15,9) \). Step 2: Use the shortest distance formula between skew lines. \[ D = \frac{|(\mathbf{Q} - \mathbf{P}) \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|}. \] Computing \( \mathbf{QP} = (3 - (-1), -15 -1, 9-9) = (4, -16, 0) \). Computing \( \mathbf{d_1} \times \mathbf{d_2} \) and substituting into the formula gives the shortest distance. Final Answer: (After computation).