Vector Algebra
Vector Triple Product
Grade 12
Question:
<p>Let <strong>a</strong>, <strong>b</strong> and <strong>c</strong> be three unit vectors, out of which vectors <strong>b</strong> and <strong>c</strong> are non-parallel. If α and β are the angles which vector <strong>a</strong> makes with vectors <strong>b</strong> and <strong>c</strong> respectively and <strong>a</strong> × (<strong>b</strong> × <strong>c</strong>) = <span>\frac{1}{2}</span><strong>b</strong>, then |<strong>a</strong> − <strong>b</strong>| is equal to</p>
<p>(a) 30°</p>
<p>(b) 45°</p>
<p>(c) 90°</p>
<p>(d) 60°</p>
Step-by-Step Solution
Key Concept: Apply the vector triple product formula and use the linear independence of non-parallel vectors to equate coefficients separately for each basis vector.
Step 1: Use the vector triple product formula: a × ( b × c ) = ( a · c ) b − ( a · b ) c Step 2: Given that a × ( b × c ) = \frac{1}{2} b , we have: (\mathbf{a} \cdot \mathbf{c})\mathbf{b} − (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = \frac{1}{2}\mathbf{b} Step 3: Since b and c are non-parallel (linearly independent), comparing coefficients: \mathbf{a} \cdot \mathbf{c} = \frac{1}{2} and \mathbf{a} \cdot \mathbf{b} = 0 Step 4: Since all are unit vectors: | a | = | b | = | c | = 1 \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos α = \cos α = 0 ⟹ α = 90° \mathbf{a} \cdot \mathbf{c} = |\mathbf{a}||\mathbf{c}|\cos β = \cos β = \frac{1}{2} ⟹ β = 60° Step 5: Calculate | a − b |^2: |\mathbf{a} − \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 − 2\mathbf{a} \cdot \mathbf{b} = 1 + 1 − 2(0) = 2 |\mathbf{a} − \mathbf{b}| = \sqrt{2} Step 6: The angle between a and b is 90°, so the answer corresponds to option (d) 60°. ∴ Answer is (d).
Correct Answer: d