let a, b, c be three vectors such that `a. (b + c) = b. (c + a) = c. (a + b) = 0 and |a| =1, |b| =4, |c| =8`, then `|a + b + c|` equals

Options

A. 13
B. 81
C. 9
D. 5

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Given Conditions
We have three vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) such that:
1. \( \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = 0 \)
2. \( \mathbf{b} \cdot (\mathbf{c} + \mathbf{a}) = 0 \)
3. \( \mathbf{c} \cdot (\mathbf{a} + \mathbf{b}) = 0 \)

Additionally, we know:
- \( |\mathbf{a}| = 1 \)
- \( |\mathbf{b}| = 4 \)
- \( |\mathbf{c}| = 8 \)

### Step 2: Expanding the Dot Products
From the first condition:
\[ \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} = 0 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{b} = -\mathbf{a} \cdot \mathbf{c} \]

From the second condition:
\[ \mathbf{b} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{a} = 0 \quad \Rightarrow \quad \mathbf{b} \cdot \mathbf{c} = -\mathbf{b} \cdot \mathbf{a} \]

From the third condition:
\[ \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} = 0 \quad \Rightarrow \quad \mathbf{c} \cdot \mathbf{a} = -\mathbf{c} \cdot \mathbf{b} \]

### Step 3: Summing the Dot Products
Now, we can sum these three equations:
\[ (\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}) + (\mathbf{b} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{a}) + (\mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b}) = 0 \]
This simplifies to:
\[ 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0 \]
Thus:
\[ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} = 0 \]

### Step 4: Finding the Magnitude of \( \mathbf{a} + \mathbf{b} + \mathbf{c} \)
We need to find \( |\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 \):
\[ |\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 = (\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) \]
Expanding this:
\[ = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) \]
Substituting the known magnitudes and the result from Step 3:
\[ = |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2(0) \]
\[ = 1^2 + 4^2 + 8^2 \]
\[ = 1 + 16 + 64 = 81 \]

### Step 5: Taking the Square Root
Finally, we take the square root to find the magnitude:
\[ |\mathbf{a} + \mathbf{b} + \mathbf{c}| = \sqrt{81} = 9 \]

### Final Answer
Thus, \( |\mathbf{a} + \mathbf{b} + \mathbf{c}| = 9 \).

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