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

Options

A. `a^(-2)`
B. `3a^(-2)`
C. `4a^(-2)`
D. `2a^(-2)`

Answer

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: Define the vector \(\vec{a}\)

Let:
\[ \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \]

### Step 2: Calculate \(\vec{a} \times \hat{i}\)

Using the determinant form for the cross product:
\[ \vec{a} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{a} \times \hat{i} = (a_2 \cdot 0 - a_3 \cdot 0) \hat{i} - (a_1 \cdot 0 - a_3 \cdot 1) \hat{j} + (a_1 \cdot 0 - a_2 \cdot 1) \hat{k} \]
\[ = -a_3 \hat{j} + a_2 \hat{k} \]
Thus,
\[ \vec{a} \times \hat{i} = -a_3 \hat{j} + a_2 \hat{k} \]

### Step 3: Calculate \(|\vec{a} \times \hat{i}|^2\)

Now, we find the magnitude squared:
\[ |\vec{a} \times \hat{i}|^2 = |-a_3 \hat{j} + a_2 \hat{k}|^2 = (-a_3)^2 + (a_2)^2 = a_3^2 + a_2^2 \]

### Step 4: Calculate \(\vec{a} \times \hat{j}\)

Now, calculate \(\vec{a} \times \hat{j}\):
\[ \vec{a} \times \hat{j} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 0 & 1 & 0 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{a} \times \hat{j} = (a_3 \cdot 0 - a_2 \cdot 0) \hat{i} - (a_1 \cdot 0 - a_3 \cdot 1) \hat{j} + (a_1 \cdot 1 - a_2 \cdot 0) \hat{k} \]
\[ = a_3 \hat{i} - a_1 \hat{k} \]

### Step 5: Calculate \(|\vec{a} \times \hat{j}|^2\)

Now, we find the magnitude squared:
\[ |\vec{a} \times \hat{j}|^2 = |a_3 \hat{i} - a_1 \hat{k}|^2 = (a_3)^2 + (-a_1)^2 = a_3^2 + a_1^2 \]

### Step 6: Calculate \(\vec{a} \times \hat{k}\)

Now, calculate \(\vec{a} \times \hat{k}\):
\[ \vec{a} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 0 & 0 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{a} \times \hat{k} = (a_2 \cdot 1 - a_3 \cdot 0) \hat{i} - (a_1 \cdot 1 - a_3 \cdot 0) \hat{j} + (a_1 \cdot 0 - a_2 \cdot 0) \hat{k} \]
\[ = a_2 \hat{i} - a_1 \hat{j} \]

### Step 7: Calculate \(|\vec{a} \times \hat{k}|^2\)

Now, we find the magnitude squared:
\[ |\vec{a} \times \hat{k}|^2 = |a_2 \hat{i} - a_1 \hat{j}|^2 = (a_2)^2 + (-a_1)^2 = a_2^2 + a_1^2 \]

### Step 8: Combine all results

Now, combine all the results:
\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_3^2 + a_1^2) + (a_2^2 + a_1^2) \]
\[ = 2a_1^2 + 2a_2^2 + 2a_3^2 \]

### Step 9: Factor out the common term

Factoring out the common term:
\[ = 2(a_1^2 + a_2^2 + a_3^2) \]

### Step 10: Relate to the modulus of \(\vec{a}\)

Since the modulus of \(\vec{a}\) is given by:
\[ |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \]
Thus, we can write:
\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \]

### Final Answer

The final answer is:
\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \]