Let `|vec(a)| # 0.|vec(b)| ne 0 (vec(a) + vec(b)). (vec(a) + vec(b)) = |vec(a)|^(2) + |vec(b)|^(2)` holds if and only if

Options

A. `vec(a) and vec(b)` are perpendicular
B. `vec(a) and vec(b)` are parallel
C. `vec(a) and vec(b)` are inclined at an angle of `45^(@)`
D. `vec(a) and vec(b)` are anti-parallel

Answer

To solve the problem, we need to analyze the given expression:

\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 \]

### Step 1: Expand the left-hand side
Using the distributive property of the dot product, we can expand the left-hand side:

\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \]

### Step 2: Simplify the expression
This simplifies to:

\[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \]

### Step 3: Set the equation
Now we set this equal to the right-hand side of the original equation:

\[ |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 \]

### Step 4: Cancel common terms
We can subtract \( |\vec{a}|^2 + |\vec{b}|^2 \) from both sides:

\[ 2(\vec{a} \cdot \vec{b}) = 0 \]

### Step 5: Solve for the dot product
This implies:

\[ \vec{a} \cdot \vec{b} = 0 \]

### Step 6: Interpret the result
The dot product of two vectors is zero if and only if the vectors are perpendicular to each other. Therefore, we conclude that:

\[ \vec{a} \text{ and } \vec{b} \text{ are perpendicular.} \]

### Final Answer
The condition holds if and only if \(\vec{a}\) and \(\vec{b}\) are perpendicular.

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