<p>Let the equation of plane <em>P</em> be <em>(2x + 3y + z + 5) + λ(x + y + z - 6) = 0</em>. Given that the plane is perpendicular to the <em>xy</em>-plane, find the distance of point (0, 0, 256) from plane <em>P</em> (rounded to 3 decimal places).</p>
Step-by-Step Solution
Key Concept: A plane is perpendicular to the xy-plane if and only if its normal vector is perpendicular to the z-axis, meaning the z-component of the normal must be zero. Use this condition to find λ, then calculate the point-to-plane distance.
Step 1: Expand the given equation: (2x + 3y + z + 5) + λ(x + y + z - 6) = 0 This gives: (2 + λ)x + (3 + λ)y + (1 + λ)z + (5 - 6λ) = 0 Step 2: For the plane to be perpendicular to the xy-plane, its normal vector (2 + λ, 3 + λ, 1 + λ) must be perpendicular to the z-axis (normal to xy-plane is (0, 0, 1)). The normal must lie in the xy-plane, so the z-component = 0: 1 + λ = 0 ⟹ λ = -1 Step 3: Substitute λ = -1: (2 - 1)x + (3 - 1)y + (5 + 6) = 0 Simplifying: x + 2y + 11 = 0 Step 4: Calculate distance from (0, 0, 256) to plane x + 2y + 11 = 0: d = |0 + 2(0) + 11|/√(1^2 + 2^2) = 11/√5 = 11√5/5 Step 5: Compute numerically: 11 × 2.236.../5 ≈ 24.597.../5 ≈ 4.919 ∴ Answer: 4.919
Correct Answer: 4.919