3D Geometry
Equation of a Plane
Grade 12
Question:
<p>Any plane through (1, 0, 0) is \(a(x-1) + by + cz = 0\). It passes through (0, 1, 0) and makes an angle of \(\frac{\pi}{4}\) with \(x + y = 3\). Find the ratio \(a : b : c\).</p>
<p>\(1 : 1 : \sqrt{2}\)</p>
<p>\(1 : 1 : \sqrt{3}\)</p>
<p>\(\sqrt{2} : 1 : 1\)</p>
<p>\(1 : \sqrt{2} : 1\)</p>
Step-by-Step Solution
Key Concept: Use the condition that the plane passes through (0,1,0) to get a linear relation between a, b, c, then apply the angle formula between two planes using their normal vectors to get another constraint.
Step 1: The plane through (1,0,0) is a(x-1) + by + cz = 0. Since it passes through (0,1,0): a(0-1) + b(1) + c(0) = 0 -a + b = 0 ⟹ b = a Step 2: The plane can be written as a(x-1) + ay + cz = 0 or ax + ay + cz = a (after shifting). Its normal vector is n_1 = (a, a, c). Step 3: The plane x + y = 3 has normal vector n_2 = (1, 1, 0). Step 4: The angle between planes satisfies: cos(π/4) = | n_1 · n_2 | / (| n_1 | | n_2 |) 1/√2 = |a(1) + a(1) + c(0)| / (√(a^2 + a^2 + c^2) · √2) 1/√2 = |2a| / (√(2a^2 + c^2) · √2) 1 = |2a| / √(2a^2 + c^2) √(2a^2 + c^2) = 2|a| 2a^2 + c^2 = 4a^2 c^2 = 2a^2 c = ±√2 a Step 5: Taking c = √2 a (positive value): a : b : c = a : a : √2a = 1 : 1 : √2 ∴ Answer: A
Correct Answer: A