If `|vecalpha + vecbeta| = |vecalpha - vecbeta|`, then

Options

A. `alpha` is parallel to `vecbeta`
B. `alpha` is perpendicular to `vecbeta`
C. `|vecalpha| = |vecbeta|`
D. none of these

Answer

To solve the problem where \( |\vec{\alpha} + \vec{\beta}| = |\vec{\alpha} - \vec{\beta}| \), we can follow these steps:

### Step 1: Square both sides of the equation
We start with the equation:
\[ |\vec{\alpha} + \vec{\beta}| = |\vec{\alpha} - \vec{\beta}| \]
Squaring both sides, we have:
\[ |\vec{\alpha} + \vec{\beta}|^2 = |\vec{\alpha} - \vec{\beta}|^2 \]

### Step 2: Expand both sides
Using the property of magnitudes, we expand both sides:
\[ |\vec{\alpha}|^2 + |\vec{\beta}|^2 + 2|\vec{\alpha}||\vec{\beta}|\cos\theta = |\vec{\alpha}|^2 + |\vec{\beta}|^2 - 2|\vec{\alpha}||\vec{\beta}|\cos\theta \]
where \(\theta\) is the angle between vectors \(\vec{\alpha}\) and \(\vec{\beta}\).

### Step 3: Simplify the equation
Now, we can simplify the equation by cancelling out common terms:
\[ |\vec{\alpha}|^2 + |\vec{\beta}|^2 + 2|\vec{\alpha}||\vec{\beta}|\cos\theta = |\vec{\alpha}|^2 + |\vec{\beta}|^2 - 2|\vec{\alpha}||\vec{\beta}|\cos\theta \]
This leads to:
\[ 2|\vec{\alpha}||\vec{\beta}|\cos\theta + 2|\vec{\alpha}||\vec{\beta}|\cos\theta = 0 \]
or
\[ 4|\vec{\alpha}||\vec{\beta}|\cos\theta = 0 \]

### Step 4: Solve for \(\cos\theta\)
From the equation \(4|\vec{\alpha}||\vec{\beta}|\cos\theta = 0\), we can conclude:
\[ \cos\theta = 0 \]
This implies that:
\[ \theta = 90^\circ \]

### Conclusion
Thus, the angle between the vectors \(\vec{\alpha}\) and \(\vec{\beta}\) is \(90^\circ\), meaning they are perpendicular to each other.

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