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:

Memorize this standard vector identity for objective tests: For any general vector \(\vec{a}\), the expansion identity \(\sum |\hat{i} \times (\vec{a} \times \hat{i})|^2\) always simplifies cleanly to \(2|\vec{a}|^2\).

Answer 1

The Correct Option is B

Concept: This question can be simplified using the Vector Triple Product vector identity: \[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} \] Let us assume a general vector \(\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}\). We can apply this triple product property to each term individually where \(\vec{x}\) and \(\vec{z}\) are replaced by the standard unit orthogonal vectors \(\hat{i}\), \(\hat{j}\), or \(\hat{k}\). Step 1: Simplifying the first vector triple product component.
Applying the rule to \(\hat{i} \times (\vec{a} \times \hat{i})\): \[ \hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} \] Since \(\hat{i} \cdot \hat{i} = 1\) and \(\hat{i} \cdot \vec{a} = x\) (the scalar \(x\)-component of vector \(\vec{a}\)): \[ \hat{i} \times (\vec{a} \times \hat{i}) = \vec{a} - x\hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) - x\hat{i} = y\hat{j} + z\hat{k} \] Finding the squared magnitude of this resulting component vector: \[ |\hat{i} \times (\vec{a} \times \hat{i})|^2 = |y\hat{j} + z\hat{k}|^2 = y^2 + z^2 \]

Step 2: Extrapolating the result for the \(\hat{j}\) and \(\hat{k}\) components.
By symmetric property distribution, executing the same vector cross operations for the \(\hat{j}\) and \(\hat{k}\) structures yields: \[ |\hat{j} \times (\vec{a} \times \hat{j})|^2 = |x\hat{i} + z\hat{k}|^2 = x^2 + z^2 \] \[ |\hat{k} \times (\vec{a} \times \hat{k})|^2 = |x\hat{i} + y\hat{j}|^2 = x^2 + y^2 \]

Step 3: Summing the three expressions together.
Add the three squared magnitude evaluations together to express the total value: \[ \text{Total Sum} = (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2(x^2 + y^2 + z^2) = 2|\vec{a}|^2 \]

Step 4: Substituting the numerical components of the given vector \(\vec{a}\).
We are given \(\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}\), which means \(x = 2\), \(y = 1\), and \(z = 2\). \[ |\vec{a}|^2 = x^2 + y^2 + z^2 = (2)^2 + (1)^2 + (2)^2 = 4 + 1 + 4 = 9 \] Substituting this value back into our total sum equation: \[ \text{Total Value} = 2(9) = 18 \]