3D Geometry
Equation of plane
Grade 12
Question:
<p>The equation of the plane containing the line \(2x - 5y + z = 3\); \(x + y + 4z = 5\), and parallel to the plane, \(x + 3y + 6z = 1\), is</p>
<p>\(x + 3y + 6z = -7\)</p>
<p>\(x + 3y + 6z = 7\)</p>
<p>\(2x + 6y + 12z = -13\)</p>
<p>\(2x + 6y + 12z = 13\)</p>
Step-by-Step Solution
Key Concept: A plane containing a given line must satisfy the line's equation as a linear combination of the two planes defining it. For the plane to be parallel to another plane, their normal vectors must be proportional.
Step 1: The family of planes containing the line of intersection of 2x - 5y + z = 3 and x + y + 4z = 5 is: λ(2x - 5y + z - 3) + μ(x + y + 4z - 5) = 0 Expanding: (2λ + μ)x + (-5λ + μ)y + (λ + 4μ)z - (3λ + 5μ) = 0 Step 2: For this plane to be parallel to x + 3y + 6z = 1, the normal vectors must be proportional: (2λ + μ, -5λ + μ, λ + 4μ) = k(1, 3, 6) Step 3: This gives us: 2λ + μ = k ... (i) -5λ + μ = 3k ... (ii) λ + 4μ = 6k ... (iii) Step 4: From (i) - (ii): 7λ = -2k, so λ = -2k/7 From (i): μ = k - 2λ = k + 4k/7 = 11k/7 Step 5: Verify with (iii): λ + 4μ = -2k/7 + 44k/7 = 42k/7 = 6k ✓ Step 6: Taking k = 7 (for simplicity): λ = -2, μ = 11 The equation becomes: -2(2x - 5y + z - 3) + 11(x + y + 4z - 5) = 0 -4x + 10y - 2z + 6 + 11x + 11y + 44z - 55 = 0 7x + 21y + 42z - 49 = 0 Dividing by 7: x + 3y + 6z = 7 ∴ Answer: D
Correct Answer: D