<p>The shortest distance between the lines \(\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}\) and \(\dfrac{x+2}{-1} = \dfrac{y-4}{8} = \dfrac{z-5}{4}\) lies in the interval</p>
Step-by-Step Solution
Key Concept: For skew lines, the shortest distance is the perpendicular distance between them. Use the formula: d = |((b₂-b₁)·(d₁×d₂))| / |d₁×d₂|, where b₁, b₂ are position vectors and d₁, d₂ are direction vectors.
Step 1: Identify the lines in parametric form: Line 1: r_1 = (0,0,0) + t(2,2,1); point b_1 = (0,0,0), direction d_1 = (2,2,1) Line 2: r_2 = (-2,4,5) + s(-1,8,4); point b_2 = (-2,4,5), direction d_2 = (-1,8,4) Step 2: Calculate the cross product d_1 × d_2 : d_1 × d_2 = | i j k | = (2·4 - 1·8) i - (2·4 - 1·(-1)) j + (2·8 - 2·(-1)) k = (8-8) i - (8+1) j + (16+2) k = (0,-9,18) | d_1 × d_2 | = √(0 + 81 + 324) = √405 = 9√5 Step 3: Calculate b_2 - b_1 = (-2,4,5) - (0,0,0) = (-2,4,5) Step 4: Find the scalar triple product: ( b_2 - b_1 ) · ( d_1 × d_2 ) = (-2,4,5) · (0,-9,18) = 0 - 36 + 90 = 54 Step 5: Apply the distance formula: d = |54| / (9√5) = 6/√5 = (6√5)/5 ≈ 2.68 ∴ The shortest distance lies in the interval containing this value (typically around (2, 3) or (6√5/5, 3))
Correct Answer: A