If `veca, vecb, vecc` are unit vectors, then `|veca-vecb|^2+|vecb-vecc|^2+|vecc^2-veca^2|^2` does not exceed (A) 4 (B) 9 (C) 8 (D) 6

Options

A. 4
B. 9
C. 8
D. 6

Answer

To solve the problem, we need to evaluate the expression \( |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \) given that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors.

### Step-by-step Solution:

1. **Expand the squares**:
We start by expanding each term using the formula \( |\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y} \).

- For \( |\vec{a} - \vec{b}|^2 \):
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} \]
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \):
\[ |\vec{a} - \vec{b}|^2 = 1 + 1 - 2\vec{a} \cdot \vec{b} = 2 - 2\vec{a} \cdot \vec{b} \]

- For \( |\vec{b} - \vec{c}|^2 \):
\[ |\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} = 1 + 1 - 2\vec{b} \cdot \vec{c} = 2 - 2\vec{b} \cdot \vec{c} \]

- For \( |\vec{c} - \vec{a}|^2 \):
\[ |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a} = 1 + 1 - 2\vec{c} \cdot \vec{a} = 2 - 2\vec{c} \cdot \vec{a} \]

2. **Combine the results**:
Now, we combine all the expanded terms:
\[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = (2 - 2\vec{a} \cdot \vec{b}) + (2 - 2\vec{b} \cdot \vec{c}) + (2 - 2\vec{c} \cdot \vec{a}) \]
Simplifying this gives:
\[ = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

3. **Analyzing the dot products**:
Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, the dot products \( \vec{a} \cdot \vec{b}, \vec{b} \cdot \vec{c}, \vec{c} \cdot \vec{a} \) can each range from -1 to 1. Therefore, the sum \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) can range from -3 to 3.

4. **Finding the maximum value**:
The maximum value of \( 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \) occurs when \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is minimized. The minimum occurs at -3:
\[ 6 - 2(-3) = 6 + 6 = 12 \]
However, this is not possible since the vectors are unit vectors. The maximum occurs when the dot products are at their maximum (which is 3):
\[ 6 - 2(3) = 6 - 6 = 0 \]

5. **Finding the minimum value**:
The minimum occurs when the dot products are at their maximum (which is 3):
\[ 6 - 2(3) = 6 - 6 = 0 \]

6. **Conclusion**:
Thus, the expression \( |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \) does not exceed 6.

### Final Answer:
The answer is (D) 6.