3D Geometry
Plane Equations
Grade 12
Question:
<p>Equation of plane passing through the points <span class="math">C_1(2, 2, 1)</span>, <span class="math">(9, 3, 6)</span> and perpendicular to the plane <span class="math">2x + 6y + 6z - 1 = 0</span>, is</p>
<p>(a) <span class="math">3x + 4y + 5z = 9</span></p>
<p>(b) <span class="math">3x + 4y - 5z + 9 = 0</span></p>
<p>(c) <span class="math">3x + 4y - 5z - 9 = 0</span></p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Use the point form of plane equation and apply the perpendicularity condition. The normal vectors of perpendicular planes are orthogonal.
Step 1: Equation of a plane passing through (2, 2, 1) is a(x - 2) + b(y - 2) + c(z - 1) = 0 Step 2: This plane passes through (9, 3, 6) , so 7a + b + 5c = 0 Step 3: The plane is perpendicular to 2x + 6y + 6z - 1 = 0 , so 2a + 6b + 6c = 0 Step 4: Solving by cross-multiplication: \frac{a}{-24} = \frac{b}{-32} = \frac{c}{40} , which gives \frac{a}{-3} = \frac{b}{-4} = \frac{c}{5} Step 5: Substituting into the plane equation: 3x + 4y - 5z - 9 = 0 ∴ Answer is (c).
Correct Answer: C