`u = a xx (b xx c) + b xx ( c xx a) + c xx ( a xx b)` , 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 can use the vector triple product identity, which states that:

\[ a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \]

We will apply this identity to each term in the expression for \( u \).

### Step 1: Apply the vector triple product identity to the first term
Using the identity on the first term \( a \times (b \times c) \):

\[ a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \]

### Step 2: Apply the vector triple product identity to the second term
Now, apply the identity to the second term \( b \times (c \times a) \):

\[ b \times (c \times a) = (b \cdot a)c - (b \cdot c)a \]

### Step 3: Apply the vector triple product identity to the third term
Next, apply the identity to the third term \( c \times (a \times b) \):

\[ c \times (a \times b) = (c \cdot b)a - (c \cdot a)b \]

### Step 4: Combine all the terms
Now, substitute all three results back into the expression for \( u \):

\[ u = [(a \cdot c)b - (a \cdot b)c] + [(b \cdot a)c - (b \cdot c)a] + [(c \cdot b)a - (c \cdot a)b] \]

### Step 5: Rearrange and simplify
Now, let's rearrange the terms:

\[ u = (a \cdot c)b - (a \cdot b)c + (b \cdot a)c - (b \cdot c)a + (c \cdot b)a - (c \cdot a)b \]

### Step 6: Group similar terms
Grouping similar terms gives:

\[ u = [(a \cdot c)b - (c \cdot a)b] + [(b \cdot a)c - (a \cdot b)c] + [(c \cdot b)a - (b \cdot c)a] \]

### Step 7: Factor out common terms
Factoring out the common terms leads to:

\[ u = (a \cdot c - c \cdot a)b + (b \cdot a - a \cdot b)c + (c \cdot b - b \cdot c)a \]

### Step 8: Recognize that dot products are commutative
Since the dot product is commutative, we have:

\[ a \cdot c - c \cdot a = 0, \quad b \cdot a - a \cdot b = 0, \quad c \cdot b - b \cdot c = 0 \]

### Conclusion
Thus, all terms cancel out, leading to:

\[ u = 0 \]

### Final Answer
The final result is:

\[ u = \mathbf{0} \]

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