<p>The perpendicular distance from the origin to the plane containing the two lines, \(\dfrac{x+2}{3} = \dfrac{y-2}{5} = \dfrac{z+5}{7}\) and \(\dfrac{x-1}{1} = \dfrac{y-4}{4} = \dfrac{z+4}{7}\), is ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: Two skew lines in 3D determine a unique plane only if they are coplanar (intersecting or parallel). First verify coplanarity using the scalar triple product condition: (P₂-P₁)·(d₁×d₂) = 0, then find the plane equation using the normal vector d₁×d₂.
Step 1: Extract line data
Line 1: Point P_1 = (-2, 2, -5), direction d_1 = (3, 5, 7)
Line 2: Point P_2 = (1, 4, -4), direction d_2 = (1, 4, 7) Step 2: Verify coplanarity
P_2 - P_1 = (3, 2, 1)
d_1 × d_2 = (5·7 - 7·4, 7·1 - 3·7, 3·4 - 5·1) = (35-28, 7-21, 12-5) = (7, -14, 7)
(P_2-P_1)·(d_1×d_2) = 3(7) + 2(-14) + 1(7) = 21 - 28 + 7 = 0 ✓ (coplanar) Step 3: Find plane equation
Normal vector: n = (7, -14, 7) = 7(1, -2, 1)
Using point P_1(-2, 2, -5):
1(x+2) - 2(y-2) + 1(z+5) = 0
x + 2 - 2y + 4 + z + 5 = 0
x - 2y + z + 11 = 0 Step 4: Calculate distance from origin
Distance = |0 - 0 + 0 + 11|/√(1^2 + (-2)^2 + 1^2) = 11/√6 = 11√6/6 ≈ 4.495
∴ Answer: 4.495
Correct Answer: 4