Vector Algebra
Angle between vectors
Grade 12
Question:
<p>\(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a}| = 1\), \(|\vec{b}| = 4\) and \(\vec{a} \cdot \vec{b} = 2\). If \(\vec{c} = \left(2\vec{a} \times \vec{b}\right) - 3\vec{b}\), then find the angle between \(\vec{b}\) and \(\vec{c}\).</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{3\pi}{4}\)</p>
<p>\(\dfrac{5\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: The angle between vectors is found using the dot product formula cos θ = (u·v)/(|u||v|). Since c involves a cross product term (perpendicular to b) and a parallel term (-3b), decompose c strategically: the cross product component is perpendicular to b, making the angle depend only on the dot product with the parallel component.
Step 1: Find sin θ from the given dot product. Given: |a| = 1, |b| = 4, a·b = 2 cos θ_0 = (a·b)/(|a||b|) = 2/(1×4) = 1/2, so θ_0 = 60° sin θ_0 = sin 60° = √3/2 Step 2: Calculate |a × b|. |a × b| = |a||b|sin θ_0 = 1 × 4 × (√3/2) = 2√3 Step 3: Analyze vector c = 2(a × b) - 3b. Since (a × b) ⊥ a and (a × b) ⊥ b, the term 2(a × b) is perpendicular to b. Therefore: (a × b) · b = 0 Step 4: Calculate b · c. b · c = b · [2(a × b) - 3b] = 2[b · (a × b)] - 3(b · b) b · c = 2(0) - 3|b|^2 = -3(16) = -48 Step 5: Calculate |c|. |c|^2 = |2(a × b) - 3b|^2 = |2(a × b)|^2 + |-3b|^2 (since they're perpendicular) |c|^2 = 4|a × b|^2 + 9|b|^2 = 4(12) + 9(16) = 48 + 144 = 192 |c| = 8√3 Step 6: Find the angle between b and c. cos φ = (b · c)/(|b||c|) = -48/(4 × 8√3) = -48/(32√3) = -3/(2√3) = -√3/2 Therefore φ = 150° ∴ Answer: D (150°)
Correct Answer: D