For any vector `vecp` , the value of
`3/2 { |vecp xx hati|^2 + |vecp xx hatj|^2 + | vecp xx hatk|^2}` is

Options

A. `vecp^2`
B. `2vecp^2`
C. `3vecp^2`
D. `4vecp^2`

Answer

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

\[ \frac{3}{2} \left( |\vec{p} \times \hat{i}|^2 + |\vec{p} \times \hat{j}|^2 + |\vec{p} \times \hat{k}|^2 \right) \]

where \(\vec{p}\) is a vector given by:

\[ \vec{p} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \]

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

Using the determinant form for the cross product, we have:

\[ \vec{p} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{vmatrix} \]

Calculating this determinant, we find:

\[ \vec{p} \times \hat{i} = (a_2 \cdot 0 - a_3 \cdot 0) \hat{i} - (a_1 \cdot 0 - a_3 \cdot 1) \hat{j} + (a_1 \cdot 0 - a_2 \cdot 1) \hat{k} = -a_3 \hat{j} + a_2 \hat{k} \]

Thus,

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

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

Now we calculate:

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

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

Similarly, for \(\vec{p} \times \hat{j}\):

\[ \vec{p} \times \hat{j} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 0 & 1 & 0 \end{vmatrix} \]

Calculating this determinant gives:

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

Thus,

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

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

Now we calculate:

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

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

For \(\vec{p} \times \hat{k}\):

\[ \vec{p} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 0 & 0 & 1 \end{vmatrix} \]

Calculating this determinant gives:

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

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

Now we calculate:

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

### Step 7: Substitute back into the original expression

Now we substitute the values into the original expression:

\[ \frac{3}{2} \left( (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \right) \]

This simplifies to:

\[ \frac{3}{2} \left( 2a_1^2 + 2a_2^2 + 2a_3^2 \right) = 3(a_1^2 + a_2^2 + a_3^2) \]

### Step 8: Final result

Since \( |\vec{p}|^2 = a_1^2 + a_2^2 + a_3^2 \), we can write:

\[ 3(a_1^2 + a_2^2 + a_3^2) = 3|\vec{p}|^2 \]

Thus, the final answer is:

\[ \frac{3}{2} \left( |\vec{p} \times \hat{i}|^2 + |\vec{p} \times \hat{j}|^2 + |\vec{p} \times \hat{k}|^2 \right) = 3|\vec{p}|^2 \]