<p>The equation of the plane that passes through the points <i>(1, 1, 0)</i>, <i>(1, 2, 1)</i>, and <i>(−2, 2, −1)</i> is</p>
<p>(a) <i>2x + 3y − 3z = 5</i></p>
<p>(b) <i>2x + 3y + 3z = 5</i></p>
<p>(c) <i>2x − 3y + 3z = 5</i></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: To find a plane through three points, we use two vectors in the plane and compute their cross product to get the normal vector. The plane equation is then formed using this normal and any point on the plane.
Step 1: Find two vectors in the plane. Let A = (1, 1, 0), B = (1, 2, 1), C = (−2, 2, −1) Vector AB = B − A = (1−1, 2−1, 1−0) = (0, 1, 1) Vector AC = C − A = (−2−1, 2−1, −1−0) = (−3, 1, −1) Step 2: Find the normal vector using cross product. n = AB × AC = | i j k | |0 1 1| |−3 1 −1| n = i (1·(−1) − 1·1) − j (0·(−1) − 1·(−3)) + k (0·1 − 1·(−3)) n = i (−1−1) − j (0+3) + k (0+3) n = −2 i − 3 j + 3 k n = (−2, −3, 3) Step 3: Write the plane equation. Using normal vector (−2, −3, 3) and point A(1, 1, 0): −2(x−1) − 3(y−1) + 3(z−0) = 0 −2x + 2 − 3y + 3 + 3z = 0 −2x − 3y + 3z + 5 = 0 2x + 3y − 3z = 5 Step 4: Verify with all three points. Point A(1, 1, 0): 2(1) + 3(1) − 3(0) = 2 + 3 = 5 ✓ Point B(1, 2, 1): 2(1) + 3(2) − 3(1) = 2 + 6 − 3 = 5 ✓ Point C(−2, 2, −1): 2(−2) + 3(2) − 3(−1) = −4 + 6 + 3 = 5 ✓ ∴ Answer: A
Correct Answer: A