Question

If \( \mathbf{b} \) and \( \mathbf{c} \) are any two non-collinear unit vectors and \( \mathbf{a} \) is any vector, then \[ (\mathbf{a} \cdot \mathbf{b})\mathbf{b} + (\mathbf{a} \cdot \mathbf{c})\mathbf{c} + \frac{(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))}{|\mathbf{b} \times \mathbf{c}|^2} (\mathbf{b} \times \mathbf{c}) \] is equal to:

• A. 0
• B. \(\mathbf{a}\)
• C. \(\mathbf{b}\)
• D. \(\mathbf{c}\)

Hint:

In vector algebra, \(\mathbf{a} = (\mathbf{a} \cdot \hat{\mathbf{i}})\hat{\mathbf{i}} + (\mathbf{a} \cdot \hat{\mathbf{j}})\hat{\mathbf{j}} + (\mathbf{a} \cdot \hat{\mathbf{k}})\hat{\mathbf{k}}\) for orthonormal basis. For non-orthogonal basis, a similar decomposition exists using reciprocal basis.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
\(\mathbf{b}, \mathbf{c}, \mathbf{b} \times \mathbf{c}\) form a basis in \(\mathbb{R}^3\).

Step 2: Detailed Explanation:
Any vector \(\mathbf{a}\) can be expressed as \(\mathbf{a} = \alpha \mathbf{b} + \beta \mathbf{c} + \gamma (\mathbf{b} \times \mathbf{c})\).
Taking dot product with \(\mathbf{b}\): \(\mathbf{a} \cdot \mathbf{b} = \alpha + \beta(\mathbf{c} \cdot \mathbf{b}) + 0\). Since \(\mathbf{b}\) and \(\mathbf{c}\) are unit and non-collinear, they are not necessarily orthogonal. But the given expression is exactly the decomposition of \(\mathbf{a}\) in the reciprocal basis. The expression \((\mathbf{a} \cdot \mathbf{b})\mathbf{b} + (\mathbf{a} \cdot \mathbf{c})\mathbf{c}\) gives the projection onto the plane of \(\mathbf{b}, \mathbf{c}\), and the third term gives the component perpendicular. The complete sum equals \(\mathbf{a}\).

Step 3: Final Answer:
Option (B) \(\mathbf{a}\).