Question

If \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \), then the value of \( |\hat{i} \times (\vec{a} \times \hat{i})|^2 + |\hat{j} \times (\vec{a} \times \hat{j})|^2 + |\hat{k} \times (\vec{a} \times \hat{k})|^2 \) is equal to

• A. 17
• B. 18
• C. 19
• D. 20

Hint:

The expression \( \sum |\hat{u} \times (\vec{a} \times \hat{u})|^2 \) for the orthonormal basis vectors is always equal to \( 2|\vec{a}|^2 \). This is a useful identity in vector algebra!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The expression involves the Vector Triple Product. The formula for a triple product is \( \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} \).

Step 2: Key Formula or Approach
For the first term: \( \hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} = \vec{a} - a_x\hat{i} \). Since \( \vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k} \), then \( \vec{a} - a_x\hat{i} = a_y\hat{j} + a_z\hat{k} \).

Step 3: Detailed Calculation
1. Magnitude of the first term: \( |a_y\hat{j} + a_z\hat{k}|^2 = a_y^2 + a_z^2 \). 2. Similarly, for the second term: \( |a_x\hat{i} + a_z\hat{k}|^2 = a_x^2 + a_z^2 \). 3. Similarly, for the third term: \( |a_x\hat{i} + a_y\hat{j}|^2 = a_x^2 + a_y^2 \). 4. Summing them up: \[ \text{Total} = (a_y^2 + a_z^2) + (a_x^2 + a_z^2) + (a_x^2 + a_y^2) = 2(a_x^2 + a_y^2 + a_z^2) = 2|\vec{a}|^2 \] 5. Given \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \), so \( |\vec{a}|^2 = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9 \). 6. Result: \( 2 \times 9 = 18 \).

Step 4: Final Answer
The value of the expression is 18.