For any vector ` veca`
` |veca xx hati|^(2) + |veca xx hatj|^(2) + |veca xx hatk|^(2) ` is equal to

Options

A. ` |veca|^(2)`
B. ` 2 |veca|^(2)`
C. ` 3|veca|^(2)`
D. `2 |veca|`

Explanation

To solve the problem, we need to evaluate the expression:

\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \]

where \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\).

### Step 1: Calculate \(\vec{a} \times \hat{i}\)

Using the definition of the cross product:

\[ \vec{a} \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} \]

Calculating each term:

- \(\hat{i} \times \hat{i} = 0\)
- \(\hat{j} \times \hat{i} = -\hat{k}\)
- \(\hat{k} \times \hat{i} = \hat{j}\)

Thus,

\[ \vec{a} \times \hat{i} = a_2(-\hat{k}) + a_3(\hat{j}) = -a_2 \hat{k} + a_3 \hat{j} \]

### Step 2: Calculate the magnitude squared of \(\vec{a} \times \hat{i}\)

\[ |\vec{a} \times \hat{i}|^2 = |-a_2 \hat{k} + a_3 \hat{j}|^2 = (-a_2)^2 + (a_3)^2 = a_2^2 + a_3^2 \]

### Step 3: Calculate \(\vec{a} \times \hat{j}\)

Now, calculate \(\vec{a} \times \hat{j}\):

\[ \vec{a} \times \hat{j} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{j} \]

Calculating each term:

- \(\hat{i} \times \hat{j} = \hat{k}\)
- \(\hat{j} \times \hat{j} = 0\)
- \(\hat{k} \times \hat{j} = -\hat{i}\)

Thus,

\[ \vec{a} \times \hat{j} = a_1 \hat{k} - a_3 \hat{i} \]

### Step 4: Calculate the magnitude squared of \(\vec{a} \times \hat{j}\)

\[ |\vec{a} \times \hat{j}|^2 = |a_1 \hat{k} - a_3 \hat{i}|^2 = (a_1)^2 + (-a_3)^2 = a_1^2 + a_3^2 \]

### Step 5: Calculate \(\vec{a} \times \hat{k}\)

Now, calculate \(\vec{a} \times \hat{k}\):

\[ \vec{a} \times \hat{k} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{k} \]

Calculating each term:

- \(\hat{i} \times \hat{k} = -\hat{j}\)
- \(\hat{j} \times \hat{k} = \hat{i}\)
- \(\hat{k} \times \hat{k} = 0\)

Thus,

\[ \vec{a} \times \hat{k} = -a_1 \hat{j} + a_2 \hat{i} \]

### Step 6: Calculate the magnitude squared of \(\vec{a} \times \hat{k}\)

\[ |\vec{a} \times \hat{k}|^2 = |-a_1 \hat{j} + a_2 \hat{i}|^2 = (-a_1)^2 + (a_2)^2 = a_1^2 + a_2^2 \]

### Step 7: Combine the results

Now, we can combine all the results:

\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \]

This simplifies to:

\[ = 2(a_1^2 + a_2^2 + a_3^2) \]

### Step 8: Final Result

Recognizing that \(a_1^2 + a_2^2 + a_3^2\) is the square of the magnitude of \(\vec{a}\):

\[ |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \]

Thus, we can express the final result as:

\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2 |\vec{a}|^2 \]