If `veca` is any vector, then `(vec a xx vec i)^2+(vec a xx vecj)^2+(veca xx vec k)^2` is equal to

Options

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

Answer

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

\[ (\vec{a} \times \vec{i})^2 + (\vec{a} \times \vec{j})^2 + (\vec{a} \times \vec{k})^2 \]

where \(\vec{a}\) is any vector expressed in terms of its components along the unit vectors \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\).

### Step 1: Express the vector \(\vec{a}\)
Let \(\vec{a} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}\).

### Step 2: Calculate \(\vec{a} \times \vec{i}\)
Using the properties of the cross product:
\[ \vec{a} \times \vec{i} = (a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}) \times \vec{i} \]
The cross product of \(\vec{i}\) with itself is zero, and using the right-hand rule:
\[ = a_2 (\vec{j} \times \vec{i}) + a_3 (\vec{k} \times \vec{i}) = a_2 (-\vec{k}) + a_3 \vec{j} = -a_2 \vec{k} + a_3 \vec{j} \]

### Step 3: Calculate \((\vec{a} \times \vec{i})^2\)
Now we find the magnitude squared:
\[ |\vec{a} \times \vec{i}|^2 = (-a_2 \vec{k} + a_3 \vec{j}) \cdot (-a_2 \vec{k} + a_3 \vec{j}) = a_2^2 + a_3^2 \]

### Step 4: Calculate \(\vec{a} \times \vec{j}\)
Next, we calculate \(\vec{a} \times \vec{j}\):
\[ \vec{a} \times \vec{j} = (a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}) \times \vec{j} \]
Using the properties of the cross product:
\[ = a_1 (\vec{i} \times \vec{j}) + a_3 (\vec{k} \times \vec{j}) = a_1 \vec{k} - a_3 \vec{i} \]

### Step 5: Calculate \((\vec{a} \times \vec{j})^2\)
Now we find the magnitude squared:
\[ |\vec{a} \times \vec{j}|^2 = (a_1 \vec{k} - a_3 \vec{i}) \cdot (a_1 \vec{k} - a_3 \vec{i}) = a_1^2 + a_3^2 \]

### Step 6: Calculate \(\vec{a} \times \vec{k}\)
Now we calculate \(\vec{a} \times \vec{k}\):
\[ \vec{a} \times \vec{k} = (a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}) \times \vec{k} \]
Using the properties of the cross product:
\[ = a_1 (\vec{i} \times \vec{k}) + a_2 (\vec{j} \times \vec{k}) = -a_1 \vec{j} + a_2 \vec{i} \]

### Step 7: Calculate \((\vec{a} \times \vec{k})^2\)
Now we find the magnitude squared:
\[ |\vec{a} \times \vec{k}|^2 = (-a_1 \vec{j} + a_2 \vec{i}) \cdot (-a_1 \vec{j} + a_2 \vec{i}) = a_1^2 + a_2^2 \]

### Step 8: Combine all results
Now we can combine all the results:
\[ (\vec{a} \times \vec{i})^2 + (\vec{a} \times \vec{j})^2 + (\vec{a} \times \vec{k})^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \]
This simplifies to:
\[ = 2a_1^2 + 2a_2^2 + 2a_3^2 = 2(a_1^2 + a_2^2 + a_3^2) \]

### Final Result
Thus, the final expression is:
\[ (\vec{a} \times \vec{i})^2 + (\vec{a} \times \vec{j})^2 + (\vec{a} \times \vec{k})^2 = 2|\vec{a}|^2 \]