If `veca` and `vecb` are two vecctors such that `abs(veca+vecb)=abs(veca-vecb)=4`, then which one of the following is correct?

Options

A. `veca` and `vecb` must be unit vector
B. `veca` must be parallel to `vecb`
C. `veca` must be perpendicular to `vecb`
D. `veca` must be equal to `vecb`

Answer

To solve the problem, we are given two vectors \(\vec{a}\) and \(\vec{b}\) such that:

\[ |\vec{a} + \vec{b}| = 4 \]
\[ |\vec{a} - \vec{b}| = 4 \]

We need to find the relationship between the vectors \(\vec{a}\) and \(\vec{b}\).

### Step 1: Square both equations

We start by squaring both sides of the equations:

\[ |\vec{a} + \vec{b}|^2 = 4^2 = 16 \]
\[ |\vec{a} - \vec{b}|^2 = 4^2 = 16 \]

### Step 2: Use the formula for the magnitude of vectors

Using the formula for the magnitude of vectors, we can expand both equations:

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

### Step 3: Set up the equations

From the first equation, we have:

\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b} = 16 \quad (1) \]

From the second equation, we have:

\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} = 16 \quad (2) \]

### Step 4: Subtract the equations

Now, we can subtract equation (2) from equation (1):

\[ (\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b}) - (\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b}) = 16 - 16 \]

This simplifies to:

\[ 4 \vec{a} \cdot \vec{b} = 0 \]

### Step 5: Conclusion

From this, we find that:

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

This means that the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other.

### Final Answer

The correct conclusion is that \(\vec{a}\) is perpendicular to \(\vec{b}\).

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