3D Geometry
Angle between line and plane
Grade 12
Question:
<p>If an angle between the line, \(\dfrac{x+1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{-2}\) and the plane, \(x - 2y - kz = 3\) is \(\cos^{-1}(2\sqrt{2}/3)\), then a value of \(k\) is ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: The angle θ between a line and plane is related to the angle φ between the line's direction vector and the plane's normal vector by θ = 90° - φ. Use sin(θ) = |cos(φ)| = |d·n|/(|d||n|) where d is the direction vector and n is the normal vector.
Step 1: Identify the direction vector of the line and normal vector of the plane. Direction vector: d = (2, 1, -2) Normal vector: n = (1, -2, -k) Step 2: The angle θ between a line and plane satisfies: sin(θ) = | d · n |/(| d || n |) Given: θ = cos⁻^1(2√2/3), so cos(θ) = 2√2/3 Therefore: sin(θ) = √(1 - 8/9) = √(1/9) = 1/3 Step 3: Calculate the dot product and magnitudes. d · n = 2(1) + 1(-2) + (-2)(-k) = 2 - 2 + 2k = 2k | d | = √(4 + 1 + 4) = √9 = 3 | n | = √(1 + 4 + k^2) = √(5 + k^2) Step 4: Apply the formula. 1/3 = |2k|/(3√(5 + k^2)) √(5 + k^2) = |2k| 5 + k^2 = 4k^2 5 = 3k^2 k^2 = 5/3 k = ±√(5/3) = ±√5/√3 = ±√15/3 Step 5: Calculate numerical value. k = ±1.291 (to three decimal places) ∴ Answer: 1.291 or -1.291
Correct Answer: 1