3D Geometry
Area of triangle in 3D
Grade 12
Question:
<p>Image of point <em>Q</em> in a plane is found using: \(\dfrac{x-0}{3} = \dfrac{y+1}{-1} = \dfrac{z+3}{4} = \dfrac{-2(1-12-2)}{9+1+16} = 1\). Given points <em>P</em>(3, -2, 1), <em>Q</em>(0, -1, -3) and <em>R</em>(3, -1, -2), find the area of triangle <em>PQR</em> (in square units, rounded to 3 decimal places).</p>
Step-by-Step Solution
Key Concept: The area of triangle PQR is half the magnitude of the cross product of vectors PQ and PR. The given reflection formula is a distraction—focus on the three given points P, Q, R to form two edge vectors and compute |PQ × PR|/2.
Step 1: Find vectors PQ and PR. PQ = Q − P = (0−3, −1−(−2), −3−1) = (−3, 1, −4) PR = R − P = (3−3, −1−(−2), −2−1) = (0, 1, −3) Step 2: Compute the cross product PQ × PR. PQ × PR = | i j k | = i (1·(−3) − (−4)·1) − j ((−3)·(−3) − (−4)·0) + k ((−3)·1 − 1·0) |−3 1 −4| |0 1 −3| = i (−3 + 4) − j (9 − 0) + k (−3 − 0) = i (1) − j (9) + k (−3) = (1, −9, −3) Step 3: Find the magnitude. |PQ × PR| = √(1^2 + (−9)^2 + (−3)^2) = √(1 + 81 + 9) = √91 ≈ 9.5394 Step 4: Calculate the area. Area = (1/2)|PQ × PR| = (1/2) × 9.5394 ≈ 4.7697 ∴ Answer: 4.770
Correct Answer: 4.770