For any vector `vec(a), |vec(a)xx hat(i)|^(2)+ |vec(a)xx hat(j)|^(2)+ |vec(a) xx hat(k)|^(2)` is equal to

Options

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

Explanation

To solve the problem, we need to find the value of the expression:

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

where \(\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}\).

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

Using the definition of the cross product, we can express \(\vec{a} \times \hat{i}\) as a determinant:

\[ \vec{a} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 0 & 0 \end{vmatrix} \]

Calculating this determinant, we have:

\[ \vec{a} \times \hat{i} = \hat{i}(0 - 0) - \hat{j}(z - 0) + \hat{k}(0 - y) = -z \hat{j} - y \hat{k} \]

Thus,

\[ \vec{a} \times \hat{i} = -z \hat{j} - y \hat{k} \]

### Step 2: Calculate the magnitude squared

Now, we find the magnitude squared of \(\vec{a} \times \hat{i}\):

\[ |\vec{a} \times \hat{i}|^2 = (-z)^2 + (-y)^2 = z^2 + y^2 \]

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

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

\[ \vec{a} \times \hat{j} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 0 & 1 & 0 \end{vmatrix} \]

Calculating this determinant, we have:

\[ \vec{a} \times \hat{j} = \hat{i}(0 - 0) - \hat{j}(0 - z) + \hat{k}(x - 0) = z \hat{i} - x \hat{k} \]

Thus,

\[ \vec{a} \times \hat{j} = z \hat{i} - x \hat{k} \]

### Step 4: Calculate the magnitude squared

Now, we find the magnitude squared of \(\vec{a} \times \hat{j}\):

\[ |\vec{a} \times \hat{j}|^2 = z^2 + (-x)^2 = z^2 + x^2 \]

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

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

\[ \vec{a} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 0 & 0 & 1 \end{vmatrix} \]

Calculating this determinant, we have:

\[ \vec{a} \times \hat{k} = \hat{i}(y - 0) - \hat{j}(x - 0) + \hat{k}(0 - 0) = y \hat{i} - x \hat{j} \]

Thus,

\[ \vec{a} \times \hat{k} = y \hat{i} - x \hat{j} \]

### Step 6: Calculate the magnitude squared

Now, we find the magnitude squared of \(\vec{a} \times \hat{k}\):

\[ |\vec{a} \times \hat{k}|^2 = y^2 + (-x)^2 = y^2 + x^2 \]

### Step 7: Combine the results

Now, we can combine all the results:

\[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (y^2 + z^2) + (z^2 + x^2) + (y^2 + x^2) \]

This simplifies to:

\[ = 2x^2 + 2y^2 + 2z^2 \]

### Step 8: Factor out the common term

We can factor out the 2:

\[ = 2(x^2 + y^2 + z^2) \]

### Step 9: Relate to the magnitude of \(\vec{a}\)

Since \(\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}\), the magnitude of \(\vec{a}\) is:

\[ |\vec{a}|^2 = x^2 + y^2 + z^2 \]

Thus, we can express our final result as:

\[ = 2|\vec{a}|^2 \]

### Final Answer

Therefore, the final answer is:

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

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