Question

The vector \((\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a}\vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k}\) is equal to:

• A. \(\vec{b} \times \vec{a}\)
• B. \(\vec{a}\)
• C. \(\vec{a} \times \vec{b}\)
• D. \(\vec{b}\)

Hint:

Each bracket is a scalar triple product; use cyclic symmetry to turn it into a component of \(\vec a\times\vec b\).

Answer 1

The Correct Option is C

\(\hat{i}\times\vec{a}\cdot\vec{b}\) is the scalar triple product \([\hat{i}\ \vec{a}\ \vec{b}]\). By cyclic symmetry, \([\hat{i}\ \vec{a}\ \vec{b}] = \hat{i}\cdot(\vec{a}\times\vec{b})\), the \(x\)-component of \(\vec{a}\times\vec{b}\).
Similarly the other two brackets equal \(\hat{j}\cdot(\vec{a}\times\vec{b})\) and \(\hat{k}\cdot(\vec{a}\times\vec{b})\).
The expression becomes \((\vec{a}\times\vec{b}\cdot\hat{i})\hat{i} + (\vec{a}\times\vec{b}\cdot\hat{j})\hat{j} + (\vec{a}\times\vec{b}\cdot\hat{k})\hat{k} = \vec{a}\times\vec{b}\), since any vector equals the sum of its components along \(\hat i,\hat j,\hat k\).
Answer: (3) \(\vec{a} \times \vec{b}\)