<p>We have the lines
\(\frac{x}{2} = \frac{y}{2} = \frac{z}{1}\) and \(\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}\).<br>The shortest distance between these two lines lies in the interval:</p>
Step-by-Step Solution
Key Concept: Use the skew lines distance formula: d = |(Ā̄ - Ā̃) · (ā̄ × ā̃)| / ||ā̄ × ā̃|| where points and direction vectors are extracted from symmetric equations.
Step 1: Extract line parameters Line 1: x/2 = y/2 = z/1 passes through P_1 = (0,0,0) with direction vector b _1 = (2,2,1) Line 2: (x+2)/(-1) = (y-4)/8 = (z-5)/4 passes through P_2 = (-2,4,5) with direction vector b _2 = (-1,8,4) Step 2: Calculate cross product b _1 × b _2 b _1 × b _2 = | i j k | = i (2·4 - 1·8) - j (2·4 - 1·(-1)) + k (2·8 - 2·(-1)) |2 2 1| |-1 8 4| = i (0) - j (9) + k (18) = (0, -9, 18) Step 3: Calculate magnitude || b _1 × b _2|| = √(0^2 + 81 + 324) = √405 = 9√5 Step 4: Calculate P_1P_2 · ( b _1 × b _2) P_1P_2 = (-2, 4, 5) (-2, 4, 5) · (0, -9, 18) = 0 - 36 + 90 = 54 Step 5: Apply distance formula d = |54| / (9√5) = 6/√5 = (6√5)/5 ≈ 2.68 ∴ Answer: C
Correct Answer: C