`veca, vecb ,vec c ` are three vectors with magnitude `|veca| = 4, |vecb| = 4, |vec c| = 2` and such that `veca` is perpendicular to `(vecb + vec c), vecb` is perpendicular to` (vec c +vec a)` and `vec c ` is perpendicualr to `(vec a + vec b) ` . It follows that `|veca + vecb+ vec c|` is equal to :

Options

A. 9
B. 6
C. 5
D. 4

Answer

To solve the problem, we will follow the given conditions and use vector properties step by step.

### Step 1: Write down the conditions based on the problem statement.
We know the following:
- \(|\vec{a}| = 4\)
- \(|\vec{b}| = 4\)
- \(|\vec{c}| = 2\)

The conditions given are:
1. \(\vec{a} \perp (\vec{b} + \vec{c})\) implies \(\vec{a} \cdot (\vec{b} + \vec{c}) = 0\)
2. \(\vec{b} \perp (\vec{c} + \vec{a})\) implies \(\vec{b} \cdot (\vec{c} + \vec{a}) = 0\)
3. \(\vec{c} \perp (\vec{a} + \vec{b})\) implies \(\vec{c} \cdot (\vec{a} + \vec{b}) = 0\)

### Step 2: Expand the dot products.
From the conditions, we can expand the dot products:
1. \(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0\) (Equation 1)
2. \(\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0\) (Equation 2)
3. \(\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0\) (Equation 3)

### Step 3: Use the magnitudes of the vectors.
We can calculate the squares of the magnitudes:
- \(|\vec{a}|^2 = 4^2 = 16\)
- \(|\vec{b}|^2 = 4^2 = 16\)
- \(|\vec{c}|^2 = 2^2 = 4\)

### Step 4: Calculate \(|\vec{a} + \vec{b} + \vec{c}|^2\).
Using the formula for the magnitude of the sum of vectors:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Substituting the known values:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 16 + 16 + 4 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 36 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

### Step 5: Substitute the values from the perpendicular conditions.
From Equations 1, 2, and 3, we know:
\[ \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \implies \vec{a} \cdot \vec{b} = -\vec{a} \cdot \vec{c} \]
\[ \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \implies \vec{b} \cdot \vec{c} = -\vec{b} \cdot \vec{a} \]
\[ \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \implies \vec{c} \cdot \vec{a} = -\vec{c} \cdot \vec{b} \]

Thus, all the dot products are zero:
\[ \vec{a} \cdot \vec{b} = 0, \quad \vec{b} \cdot \vec{c} = 0, \quad \vec{c} \cdot \vec{a} = 0 \]

### Step 6: Substitute back into the equation.
Now substituting these values into our equation:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 36 + 2(0) = 36 \]

### Step 7: Take the square root to find the magnitude.
\[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{36} = 6 \]

### Final Answer:
Thus, \(|\vec{a} + \vec{b} + \vec{c}| = 6\).

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