If `veca,vecb,vecc ` are perpendicular to `vecb+vecc,vecc+veca and veca+vecb` respectively and if `|veca+vecb|=6,|vecb+vecc|=8 and |vecc+veca|=10,then |veca+vecb+vecc|` (A) `5sqrt(2)` (B) 50 (C) `10sqrt(2)` (D) 10

Options

A. `5sqrt5`
B. 50
C. `10sqrt2`
D. 10

Answer

To solve the problem, we need to find the magnitude of the vector sum \(|\vec{a} + \vec{b} + \vec{c}|\) given that the vectors \(\vec{a}, \vec{b}, \vec{c}\) are perpendicular to certain combinations of each other and their magnitudes.

### Step-by-Step Solution:

1. **Understanding the Perpendicularity Conditions**:
- We have the following conditions based on the problem statement:
- \(\vec{a} \perp (\vec{b} + \vec{c})\) implies \(\vec{a} \cdot (\vec{b} + \vec{c}) = 0\) → \(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0\) (Equation 1)
- \(\vec{b} \perp (\vec{c} + \vec{a})\) implies \(\vec{b} \cdot (\vec{c} + \vec{a}) = 0\) → \(\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0\) (Equation 2)
- \(\vec{c} \perp (\vec{a} + \vec{b})\) implies \(\vec{c} \cdot (\vec{a} + \vec{b}) = 0\) → \(\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0\) (Equation 3)

2. **Using the Magnitudes**:
- We are given:
- \(|\vec{a} + \vec{b}| = 6\) → \((\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 36\) → \(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 36\) (Equation 4)
- \(|\vec{b} + \vec{c}| = 8\) → \((\vec{b} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 64\) → \(|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c} = 64\) (Equation 5)
- \(|\vec{c} + \vec{a}| = 10\) → \((\vec{c} + \vec{a}) \cdot (\vec{c} + \vec{a}) = 100\) → \(|\vec{c}|^2 + |\vec{a}|^2 + 2\vec{c} \cdot \vec{a} = 100\) (Equation 6)

3. **Adding the Equations**:
- We add Equations 4, 5, and 6:
\[ (|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 + 2\vec{c} \cdot \vec{a}) = 36 + 64 + 100 \]
- This simplifies to:
\[ 2(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 200 \]
- Dividing by 2 gives:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 100 \quad (Equation 7) \]

4. **Substituting the Dot Products**:
- From Equation 1, we know that \(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0\), thus \(\vec{a} \cdot \vec{b} = -\vec{a} \cdot \vec{c}\).
- Similarly, from Equation 2 and Equation 3, we can express the dot products in terms of each other.
- Substituting these into Equation 7 leads to:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 100 \]

5. **Finding \(|\vec{a} + \vec{b} + \vec{c}|\)**:
- Now, we can find \(|\vec{a} + \vec{b} + \vec{c}|\):
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
- From the previous equations, we know that the dot product terms sum to zero:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 100 \]
- Thus, \(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{100} = 10\).

### Final Answer:
\[ |\vec{a} + \vec{b} + \vec{c}| = 10 \]