Question

\((\mathbf{a} \cdot \hat{\mathbf{i}})(\mathbf{a} \times \hat{\mathbf{i}}) + (\mathbf{a} \cdot \hat{\mathbf{j}})(\mathbf{a} \times \hat{\mathbf{j}}) + (\mathbf{a} \cdot \hat{\mathbf{k}})(\mathbf{a} \times \hat{\mathbf{k}})\) is equal to

• A. \(3\mathbf{a}\)
• B. \(\mathbf{a}\)
• C. 0
• D. None of these

Hint:

Vector triple products often simplify to zero due to symmetry.

Answer 1

The Correct Option is C

To solve the given expression, we need to evaluate: \((\mathbf{a} \cdot \hat{\mathbf{i}})(\mathbf{a} \times \hat{\mathbf{i}}) + (\mathbf{a} \cdot \hat{\mathbf{j}})(\mathbf{a} \times \hat{\mathbf{j}}) + (\mathbf{a} \cdot \hat{\mathbf{k}})(\mathbf{a} \times \hat{\mathbf{k}})\).

We will examine each term individually:

1. Let's take an arbitrary vector \(\mathbf{a} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}}\).
2. The first term is \((\mathbf{a} \cdot \hat{\mathbf{i}})(\mathbf{a} \times \hat{\mathbf{i}})\).
   • The dot product, \(\mathbf{a} \cdot \hat{\mathbf{i}} = a_1\).
   • The cross product, \(\mathbf{a} \times \hat{\mathbf{i}} = (a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}}) \times \hat{\mathbf{i}} = a_2\hat{\mathbf{k}} - a_3\hat{\mathbf{j}}\).
   • Therefore, \((\mathbf{a} \cdot \hat{\mathbf{i}})(\mathbf{a} \times \hat{\mathbf{i}}) = a_1(a_2\hat{\mathbf{k}} - a_3\hat{\mathbf{j}}) = a_1a_2\hat{\mathbf{k}} - a_1a_3\hat{\mathbf{j}}\).
3. The second term is \((\mathbf{a} \cdot \hat{\mathbf{j}})(\mathbf{a} \times \hat{\mathbf{j}})\).
   • The dot product, \(\mathbf{a} \cdot \hat{\mathbf{j}} = a_2\).
   • The cross product, \(\mathbf{a} \times \hat{\mathbf{j}} = (a_1\hat{\mathbf{i}} + a_3\hat{\mathbf{k}}) \times \hat{\mathbf{j}} = a_3\hat{\mathbf{i}} - a_1\hat{\mathbf{k}}\).
   • Therefore, \((\mathbf{a} \cdot \hat{\mathbf{j}})(\mathbf{a} \times \hat{\mathbf{j}}) = a_2(a_3\hat{\mathbf{i}} - a_1\hat{\mathbf{k}}) = a_2a_3\hat{\mathbf{i}} - a_1a_2\hat{\mathbf{k}}\).
4. The third term is \((\mathbf{a} \cdot \hat{\mathbf{k}})(\mathbf{a} \times \hat{\mathbf{k}})\).
   • The dot product, \(\mathbf{a} \cdot \hat{\mathbf{k}} = a_3\).
   • The cross product, \(\mathbf{a} \times \hat{\mathbf{k}} = (a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}}) \times \hat{\mathbf{k}} = a_1\hat{\mathbf{j}} - a_2\hat{\mathbf{i}}\).
   • Therefore, \((\mathbf{a} \cdot \hat{\mathbf{k}})(\mathbf{a} \times \hat{\mathbf{k}}) = a_3(a_1\hat{\mathbf{j}} - a_2\hat{\mathbf{i}}) = a_3a_1\hat{\mathbf{j}} - a_2a_3\hat{\mathbf{i}}\).
5. Adding all the terms:
   • \(a_1a_2\hat{\mathbf{k}} - a_1a_3\hat{\mathbf{j}} + a_2a_3\hat{\mathbf{i}} - a_1a_2\hat{\mathbf{k}} + a_3a_1\hat{\mathbf{j}} - a_2a_3\hat{\mathbf{i}} = 0\).
   • Adding them, all the terms cancel each other out, thus the sum is \(0\).

Hence, the result of the given expression is: 0.