`u = a xx (b xx c) + b xx (c xx a) + c xx (a xxb)` then

Options

A. u is unit vector
B. `u = a + b +c `
C. u = 0
D. `u ne 0`

Answer

To solve the vector equation \( u = a \times (b \times c) + b \times (c \times a) + c \times (a \times b) \), we will use the vector triple product identity, which states that:

\[ x \times (y \times z) = (x \cdot z) y - (x \cdot y) z \]

### Step-by-Step Solution:

1. **Apply the Vector Triple Product Identity to the First Term:**
\[ a \times (b \times c) = (a \cdot c) b - (a \cdot b) c \]
So, the first term becomes:
\[ a \times (b \times c) = (a \cdot c) b - (a \cdot b) c \]

2. **Apply the Vector Triple Product Identity to the Second Term:**
\[ b \times (c \times a) = (b \cdot a) c - (b \cdot c) a \]
Thus, the second term becomes:
\[ b \times (c \times a) = (b \cdot a) c - (b \cdot c) a \]

3. **Apply the Vector Triple Product Identity to the Third Term:**
\[ c \times (a \times b) = (c \cdot b) a - (c \cdot a) b \]
Therefore, the third term becomes:
\[ c \times (a \times b) = (c \cdot b) a - (c \cdot a) b \]

4. **Combine All the Terms:**
Now, substituting all three terms back into the equation for \( u \):
\[ u = [(a \cdot c) b - (a \cdot b) c] + [(b \cdot a) c - (b \cdot c) a] + [(c \cdot b) a - (c \cdot a) b] \]

5. **Rearranging the Terms:**
Grouping similar terms together:
\[ u = (a \cdot c) b + (b \cdot a) c + (c \cdot b) a - [(a \cdot b) c + (b \cdot c) a + (c \cdot a) b] \]

6. **Observing Cancellation:**
Notice that:
- The terms \( (a \cdot c) b \) and \( (c \cdot a) b \) will cancel out.
- The terms \( (b \cdot a) c \) and \( (a \cdot b) c \) will cancel out.
- The terms \( (b \cdot c) a \) and \( (c \cdot b) a \) will cancel out.

Therefore, all terms cancel out, leading to:
\[ u = 0 \]

### Final Result:
\[ u = 0 \]