`veca,vecb,vecc` are three non-coplanar such that `veca + vecb + vecc = alpha vecd` and `vecb + vecc + vecd = beta veca`, then `veca + vecb + vecc + vecd` is equal to:

Options

A. `alphaveca`
B. `betavecb`
C. 0
D. `(alpha+beta)vecc`

Answer

To solve the problem step by step, we start with the given equations involving vectors:

1. **Given Equations:**
\[ \vec{a} + \vec{b} + \vec{c} = \alpha \vec{d} \quad \text{(1)} \]
\[ \vec{b} + \vec{c} + \vec{d} = \beta \vec{a} \quad \text{(2)} \]

2. **Expressing \(\vec{d}\) from Equation (1):**
From equation (1), we can express \(\vec{d}\) as:
\[ \vec{d} = \frac{\vec{a} + \vec{b} + \vec{c}}{\alpha} \quad \text{(3)} \]

3. **Substituting \(\vec{d}\) in Equation (2):**
Substitute equation (3) into equation (2):
\[ \vec{b} + \vec{c} + \frac{\vec{a} + \vec{b} + \vec{c}}{\alpha} = \beta \vec{a} \]

4. **Combining Terms:**
Rearranging the left-hand side:
\[ \vec{b} + \vec{c} + \frac{\vec{a}}{\alpha} + \frac{\vec{b}}{\alpha} + \frac{\vec{c}}{\alpha} = \beta \vec{a} \]
This simplifies to:
\[ \left(1 + \frac{1}{\alpha}\right) \vec{b} + \left(1 + \frac{1}{\alpha}\right) \vec{c} + \frac{1}{\alpha} \vec{a} = \beta \vec{a} \]

5. **Grouping Coefficients:**
We can express this as:
\[ (1 + \frac{1}{\alpha}) \vec{b} + (1 + \frac{1}{\alpha}) \vec{c} + \left(\frac{1}{\alpha} - \beta\right) \vec{a} = 0 \]

6. **Setting Coefficients to Zero:**
Since \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar, the coefficients must equal zero:
\[ 1 + \frac{1}{\alpha} = 0 \quad \text{(4)} \]
\[ \frac{1}{\alpha} - \beta = 0 \quad \text{(5)} \]

7. **Solving for \(\alpha\):**
From equation (4):
\[ \frac{1}{\alpha} = -1 \implies \alpha = -1 \]

8. **Finding \(\beta\):**
Substitute \(\alpha = -1\) into equation (5):
\[ \frac{1}{-1} - \beta = 0 \implies -1 - \beta = 0 \implies \beta = -1 \]

9. **Substituting Back:**
Substitute \(\alpha\) back into equation (1):
\[ \vec{a} + \vec{b} + \vec{c} = -\vec{d} \]
Thus,
\[ \vec{a} + \vec{b} + \vec{c} + \vec{d} = 0 \]

10. **Final Result:**
Therefore, we conclude that:
\[ \vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0} \]