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

To solve the given problem where we need to find the expression equivalent to: \((\mathbf{a} \cdot \mathbf{b})\mathbf{b} + (\mathbf{a} \cdot \mathbf{c})\mathbf{c} + \frac{(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))}{|\mathbf{b} \times \mathbf{c}|^2} (\mathbf{b} \times \mathbf{c})\), let's analyze the components involved.

Step 1: Understanding vector decomposition

If \(\mathbf{b}\) and \(\mathbf{c}\) are any two non-collinear unit vectors, the expression attempts to decompose vector \(\mathbf{a}\) into components parallel to \(\mathbf{b}\), \(\mathbf{c}\), and the plane perpendicular to them. Let's break it down:

• The term \((\mathbf{a} \cdot \mathbf{b})\mathbf{b}\) projects \(\mathbf{a}\) onto \(\mathbf{b}\).
• The term \((\mathbf{a} \cdot \mathbf{c})\mathbf{c}\) projects \(\mathbf{a}\) onto \(\mathbf{c}\).
• The last term, involving the cross product \(\mathbf{b} \times \mathbf{c}\), gives the component of \(\mathbf{a}\) perpendicular to both \(\mathbf{b}\) and \(\mathbf{c}\). The scalar factor \(\frac{(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))}{|\mathbf{b} \times \mathbf{c}|^2}\) ensures the term has the correct magnitude.

Step 2: Conclusion

The entire expression sums up to reconstruct the original vector \(\mathbf{a}\). This is because we are summing up its decomposed components along the basis defined by \(\mathbf{b}\), \(\mathbf{c}\), and \(\mathbf{b} \times \mathbf{c}\). These vectors essentially form a basis for the space, covering any possible vector in that space.

Hence, the given expression simplifies to:

\(\mathbf{a}\)

Thus, the correct answer is \(<\mathbf{a}\).