If `vecA + vecB = vecC` and `absA^2 + absB^2 = absC^2` , then angle between vectors `vecA` and `vecB` would be

Options

A. 0°
B. 45°
C. 90°
D. 180°

Answer

To solve the problem, we start with the equations given:

1. \(\vec{A} + \vec{B} = \vec{C}\)
2. \(|\vec{A}|^2 + |\vec{B}|^2 = |\vec{C}|^2\)

We need to find the angle between vectors \(\vec{A}\) and \(\vec{B}\).

### Step 1: Square the first equation
We square both sides of the first equation:
\[ |\vec{A} + \vec{B}|^2 = |\vec{C}|^2 \]

### Step 2: Expand the left side using the formula for the square of a sum
Using the formula \((\vec{A} + \vec{B})^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A} \cdot \vec{B}\), we can rewrite the left side:
\[ |\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A} \cdot \vec{B} = |\vec{C}|^2 \]

### Step 3: Substitute the second equation
From the second equation, we know that:
\[ |\vec{A}|^2 + |\vec{B}|^2 = |\vec{C}|^2 \]
We can substitute this into our expanded equation:
\[ |\vec{C}|^2 + 2\vec{A} \cdot \vec{B} = |\vec{C}|^2 \]

### Step 4: Simplify the equation
Subtract \(|\vec{C}|^2\) from both sides:
\[ 2\vec{A} \cdot \vec{B} = 0 \]

### Step 5: Solve for the dot product
The dot product \(\vec{A} \cdot \vec{B}\) can be expressed in terms of the magnitudes of the vectors and the cosine of the angle \(\theta\) between them:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Setting this equal to zero gives:
\[ |\vec{A}| |\vec{B}| \cos \theta = 0 \]

### Step 6: Analyze the equation
Since \(|\vec{A}|\) and \(|\vec{B}|\) are non-zero (as they are vectors), we have:
\[ \cos \theta = 0 \]

### Step 7: Determine the angle
The cosine of the angle is zero when:
\[ \theta = 90^\circ \]

Thus, the angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

### Final Answer
The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

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