<p>Let \(P \equiv (3, 4, a)\) and \(Q \equiv (3, 4, \lambda + a)\) be two points. The area of triangle \(OPQ\) (where \(O\) is the origin) is \(\frac{5}{4}[(a+1)^2 + 5]\). If the least area is \(\frac{p}{q}\) (in lowest terms), find \(p + q\).</p>
Step-by-Step Solution
Key Concept: The area of triangle OPQ depends on |OP × OQ|. Since P and Q differ only in z-coordinate by λ, the cross product magnitude is λ||(3,4,0)|| = 5λ. Minimizing the given expression by finding dA/da = 0 gives the critical point, then verify it's a minimum.
Step 1: Set up vectors from origin. OP = (3, 4, a) and OQ = (3, 4, λ + a) Step 2: Calculate cross product OP × OQ. OP × OQ = |i j k| |3 4 a| |3 4 λ+a| = i[4(λ+a) - 4a] - j[3(λ+a) - 3a] + k[12 - 12] = i(4λ) - j(3λ) + 0 = (4λ, -3λ, 0) Step 3: Find magnitude of cross product. |OP × OQ| = √(16λ^2 + 9λ^2) = √(25λ^2) = 5|λ| Step 4: Area of triangle OPQ. Area = ½|OP × OQ| = 5|λ|/2 Step 5: Match with given expression. 5|λ|/2 = (5/4)[(a+1)^2 + 5] |λ| = (1/2)[(a+1)^2 + 5] Step 6: Minimize the area expression. The expression (a+1)^2 + 5 is minimized when a = -1. Minimum value = 0 + 5 = 5 Step 7: Calculate minimum area. Minimum Area = (5/4) × 5 = 25/4 Therefore p = 25, q = 4, and p + q = 29
Correct Answer: 29