If `veca, vecb, vecc` are three non-zero vectors, then which of the following statement(s) is/are true?

Options

A. `atimes(btimesc), btimes(ctimesa), ctimes(atimesb)` form a right handed system
B. `c, (atimesb)times, atimesb` form a right handed system
C. `a*b+b*c+c*alt0, if a+b+c=0`
D. `((atimesb)*(btimesc))/((btimesc)*(atimesc))=-1, if a+b+c=0`

Answer

To determine which statements about the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are true, we will analyze each option step by step.

### Step 1: Analyze Option 1
**Statement:** \(\vec{a} \times (\vec{b} \times \vec{c})\) and \(\vec{b} \times (\vec{c} \times \vec{a})\) form a right-handed system.

**Solution:**
Using the vector triple product identity, we have:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]
\[ \vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} \]
Adding these two results does not yield a non-zero vector, indicating that they are coplanar, not forming a right-handed system.

**Conclusion:** Option 1 is **false**.

### Step 2: Analyze Option 2
**Statement:** \(\vec{c}\), \(\vec{a} \times \vec{b}\), and \(\vec{a}\) form a right-handed system.

**Solution:**
To check if these vectors form a right-handed system, we can check the scalar triple product:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) \]
If this product is non-zero, the vectors are not coplanar and form a right-handed system. Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are non-zero vectors, this scalar triple product is likely non-zero.

**Conclusion:** Option 2 is **true**.

### Step 3: Analyze Option 3
**Statement:** If \(\vec{a} + \vec{b} + \vec{c} = 0\), then \(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} < 0\).

**Solution:**
From the equation \(\vec{a} + \vec{b} + \vec{c} = 0\), we can square both sides:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \]
Expanding, we get:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
This implies:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{1}{2}(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) < 0 \]
Thus, the statement is true.

**Conclusion:** Option 3 is **true**.

### Step 4: Analyze Option 4
**Statement:** If \(\vec{a} + \vec{b} + \vec{c} = 0\), then \(\frac{\vec{a} \times \vec{b} \cdot \vec{b} \times \vec{c}}{\vec{b} \times \vec{c} \cdot \vec{a} \times \vec{c}} = -1\).

**Solution:**
Using the properties of cross products, we know:
\[ \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \]
\[ \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \]
Thus, we can simplify the left-hand side:
\[ \frac{|\vec{a} \times \vec{b}|^2}{|\vec{b} \times \vec{c}|^2} \]
Since \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{c}\) are equal in magnitude, we find that this expression evaluates to \(-1\).

**Conclusion:** Option 4 is **true**.

### Final Results:
- **Option 1:** False
- **Option 2:** True
- **Option 3:** True
- **Option 4:** True