If `veca,vecb" and "vecc` are three vectors such that `veca+vecb=vecc` and further `a^(2)+b^(2)=c^(2)`, find the angle between them `veca" and "vecb` ?

Answer

To find the angle between the vectors \(\vec{a}\) and \(\vec{b}\) given the conditions \(\vec{a} + \vec{b} = \vec{c}\) and \(a^2 + b^2 = c^2\), we can follow these steps:

### Step 1: Understand the given equations
We have two equations:
1. \(\vec{a} + \vec{b} = \vec{c}\)
2. \(a^2 + b^2 = c^2\)

### Step 2: Take magnitudes
Taking the magnitude of both sides of the first equation:
\[ |\vec{a} + \vec{b}| = |\vec{c}| \]

### Step 3: Use the formula for the magnitude of the sum of two vectors
The magnitude of the sum of two vectors can be expressed using the formula:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

### Step 4: Substitute the magnitudes
Substituting the magnitudes into the equation:
\[ |\vec{c}|^2 = a^2 + b^2 + 2ab \cos \theta \]

### Step 5: Use the second equation
From the second equation, we know:
\[ c^2 = a^2 + b^2 \]
Substituting this into the equation from Step 4 gives:
\[ c^2 = a^2 + b^2 + 2ab \cos \theta \]

### Step 6: Set the equations equal
Since both sides equal \(c^2\), we can set them equal:
\[ a^2 + b^2 = a^2 + b^2 + 2ab \cos \theta \]

### Step 7: Simplify the equation
Subtract \(a^2 + b^2\) from both sides:
\[ 0 = 2ab \cos \theta \]

### Step 8: Solve for \(\cos \theta\)
Since \(a\) and \(b\) cannot be zero, we can divide both sides by \(2ab\):
\[ \cos \theta = 0 \]

### Step 9: Find the angle \(\theta\)
If \(\cos \theta = 0\), then:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the vectors \(\vec{a}\) and \(\vec{b}\) is \(90^\circ\).

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