Vector Algebra
Vector triple product
Grade 12
Question:
<p>Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be non-zero vectors such that \((\vec{a}\times\vec{b})\times\vec{c} = \dfrac{1}{3}|\vec{b}||\vec{c}|\vec{a}\). If \(\theta\) is the acute angle between the vectors \(\vec{b}\) and \(\vec{c}\), then \(\sin\theta\) equals</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{\sqrt{2}}{3}\)</p>
<p>\(\dfrac{2}{3}\)</p>
<p>\(\dfrac{2\sqrt{2}}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the vector triple product formula (A×B)×C = (A·C)B - (B·C)A and equate coefficients to establish relationships between the magnitudes and angle. The condition forces specific geometric constraints that determine sin θ uniquely.
Step 1: Apply the vector triple product formula: (→a×→b)×→c = (→a·→c)→b - (→b·→c)→a Step 2: Given that this equals (1/3)|→b||→c|→a, we have: (→a·→c)→b - (→b·→c)→a = (1/3)|→b||→c|→a Step 3: Comparing coefficients (since →a and →b are non-zero and non-parallel for a meaningful configuration): • Coefficient of →b: (→a·→c) = 0, meaning →a ⊥ →c • Coefficient of →a: -(→b·→c) = (1/3)|→b||→c| Step 4: From the second equation: -|→b||→c|cos θ = (1/3)|→b||→c| Therefore: cos θ = -1/3 Step 5: Since θ is acute (0 < θ < π/2), we need |cos θ|: |cos θ| = 1/3, so sin θ = √(1 - cos^2θ) = √(1 - 1/9) = √(8/9) = 2√2/3 ∴ Answer: D
Correct Answer: D