Vector Algebra
Vector Triple Product
Grade 12
Question:
<p>Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be three non-coplanar vectors and \(\vec{r}\) be any arbitrary vector. Then \(\left(\vec{a} \times \vec{b}\right) \times \left(\vec{r} \times \vec{c}\right) + \left(\vec{b} \times \vec{c}\right) \times \left(\vec{r} \times \vec{a}\right) + \left(\vec{c} \times \vec{a}\right) \times \left(\vec{r} \times \vec{b}\right)\) is always equal to</p>
<p>\(\left[\vec{a}\,\vec{b}\,\vec{c}\right]\vec{r}\)</p>
<p>\(2\left[\vec{a}\,\vec{b}\,\vec{c}\right]\vec{r}\)</p>
<p>\(3\left[\vec{a}\,\vec{b}\,\vec{c}\right]\vec{r}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Use the vector triple product identity (A × B) × C = B(A·C) - A(B·C) repeatedly, then recognize that the cyclic sum telescopes to zero due to the Jacobi identity and properties of scalar triple products.
Step 1: Apply the vector triple product formula (A × B) × C = B(A·C) - A(B·C) to each term. For first term: (ā × b̄) × (r̄ × c̄) = b̄(ā·(r̄ × c̄)) - ā(b̄·(r̄ × c̄)) Using cyclic property of scalar triple product: ā·(r̄ × c̄) = r̄·(c̄ × ā) and b̄·(r̄ × c̄) = r̄·(c̄ × b̄) So first term = b̄(r̄·(c̄ × ā)) - ā(r̄·(c̄ × b̄)) Step 2: Similarly expand the second term: (b̄ × c̄) × (r̄ × ā) = c̄(r̄·(ā × b̄)) - b̄(r̄·(ā × c̄)) Step 3: Expand the third term: (c̄ × ā) × (r̄ × b̄) = ā(r̄·(b̄ × c̄)) - c̄(r̄·(b̄ × ā)) Step 4: Combine all three terms and group by vectors ā, b̄, c̄. Notice that coefficients of each vector involve cyclic combinations of scalar triple products that sum to zero by the Jacobi identity. Step 5: The terms cancel systematically: each of ā, b̄, c̄ appears with coefficients that sum to zero due to cyclic symmetry. ∴ Answer: 0 (or null vector)
Correct Answer: B