Vector Algebra
Angle Between Vectors
Grade 12
Question:
<p>Let <strong>a</strong> and <strong>b</strong> be two unit vectors. If the vectors <strong>c</strong> = <strong>a</strong> + 2<strong>b</strong> and <strong>d</strong> = 5<strong>a</strong> − 4<strong>a</strong> are perpendicular to each other, then the angle between <strong>a</strong> and <strong>b</strong> is</p>
<p>(a) \(\frac{\pi}{6}\)</p>
<p>(b) \(\frac{\pi}{2}\)</p>
<p>(c) \(\frac{\pi}{3}\)</p>
<p>(d) \(\frac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: Two perpendicular vectors have a dot product of zero. Use this condition along with the properties of unit vectors (|a| = |b| = 1) to find the angle between them.
Step 1: Identify given information • a and b are unit vectors: |a| = 1, |b| = 1 • c = a + 2b • d = 5a - 4b (assuming the problem meant this, not 5a - 4a) • c ⊥ d, so c · d = 0 Step 2: Set up the perpendicularity condition c · d = 0 (a + 2b) · (5a - 4b) = 0 Step 3: Expand the dot product (a + 2b) · (5a - 4b) = a·(5a) + a·(-4b) + 2b·(5a) + 2b·(-4b) = 5(a·a) - 4(a·b) + 10(b·a) - 8(b·b) Step 4: Use properties of unit vectors Since |a| = 1: a·a = 1 Since |b| = 1: b·b = 1 Since a·b = b·a, let a·b = cos θ (where θ is the angle between a and b) = 5(1) - 4(cos θ) + 10(cos θ) - 8(1) = 5 - 4cos θ + 10cos θ - 8 = -3 + 6cos θ Step 5: Solve for the angle -3 + 6cos θ = 0 6cos θ = 3 cos θ = 1/2 θ = π/3 ∴ Answer: C
Correct Answer: C