<p>Let a line parallel to \(z\)-axis passing through a point \(P(3, 4, a)\) intersects the plane \(x - 2y + 2z = a^2 + 4a + 1\) at \(Q\) where \(a \in R\). If least area of \(\triangle OPQ\) is equal to \(\left(\dfrac{p}{q}\right)\) where \(p\) and \(q\) are co-prime numbers, then find the value of \((p + q)\).</p>
Step-by-Step Solution
Key Concept: A line parallel to the z-axis has parametric form (3, 4, t). Find where it intersects the plane to get Q, then minimize the area of triangle OPQ using the distance formula and cross product method.
Step 1: Set up the line parallel to z-axis through P(3, 4, a). Parametric form: (x, y, z) = (3, 4, a + t) where t ∈ ℝ. Step 2: Find intersection Q with plane x - 2y + 2z = a^2 + 4a + 1. Substitute: 3 - 2(4) + 2(a + t) = a^2 + 4a + 1. This gives: 3 - 8 + 2a + 2t = a^2 + 4a + 1, so 2t = a^2 + 2a + 6, thus t = (a^2 + 2a + 6)/2. Step 3: Point Q has coordinates (3, 4, a + (a^2 + 2a + 6)/2) = (3, 4, (a^2 + 4a + 6)/2). Step 4: Find vector PQ: PQ = (0, 0, (a^2 + 4a + 6)/2 - a) = (0, 0, (a^2 + 2a + 6)/2). Step 5: Find vector OP: OP = (3, 4, a). Step 6: Calculate cross product OP × PQ. Since PQ is along z-axis direction: OP × PQ = |i j k| |3 4 a| |0 0 (a^2+2a+6)/2| = i·4·(a^2+2a+6)/2 - j·3·(a^2+2a+6)/2 = ((a^2+2a+6)/2)(4i - 3j). Step 7: Magnitude |OP × PQ| = ((a^2+2a+6)/2)·√(16+9) = ((a^2+2a+6)/2)·5 = 5(a^2+2a+6)/2. Step 8: Area of triangle OPQ = (1/2)|OP × PQ| = (1/2)·5(a^2+2a+6)/2 = 5(a^2+2a+6)/4. Step 9: Minimize f(a) = a^2 + 2a + 6. Taking derivative: f'(a) = 2a + 2 = 0, so a = -1. Step 10: At a = -1: f(-1) = 1 - 2 + 6 = 5. Minimum area = 5·5/4 = 25/4. Step 11: Here p = 25 and q = 4 are coprime (gcd(25,4) = 1). Therefore p + q = 25 + 4 = 29. ∴ Answer: 29
Correct Answer: 29