Understanding the Angle Between Vectors

The question asks us to find the angle between two vectors, \(\overrightarrow a\) and \(\overrightarrow b\), given a specific condition relating the magnitudes of their sum and difference. The condition is \( \left| {\overrightarrow a + \overrightarrow b } \right| = \left| {\overrightarrow a - \overrightarrow b } \right| \). We need to determine the angle between vectors \(\overrightarrow a\) and \(\overrightarrow b\) that satisfies this equation.

Let \(\theta\) be the angle between the vectors \(\overrightarrow a\) and \(\overrightarrow b\). We know that the magnitude squared of a vector is the dot product of the vector with itself, i.e., \(|\overrightarrow v|^2 = \overrightarrow v \cdot \overrightarrow v\). Using this property, we can square both sides of the given equation:

\( \left| {\overrightarrow a + \overrightarrow b } \right|^2 = \left| {\overrightarrow a - \overrightarrow b } \right|^2 \)

This expands to:

\( (\overrightarrow a + \overrightarrow b) \cdot (\overrightarrow a + \overrightarrow b) = (\overrightarrow a - \overrightarrow b) \cdot (\overrightarrow a - \overrightarrow b) \)

Using the distributive property of the dot product \((\overrightarrow x + \overrightarrow y) \cdot (\overrightarrow z + \overrightarrow w) = \overrightarrow x \cdot \overrightarrow z + \overrightarrow x \cdot \overrightarrow w + \overrightarrow y \cdot \overrightarrow z + \overrightarrow y \cdot \overrightarrow w\) and remembering that the dot product is commutative \((\overrightarrow a \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow a)\), we expand both sides:

Left side (Magnitude of vector sum):

\( (\overrightarrow a + \overrightarrow b) \cdot (\overrightarrow a + \overrightarrow b) = \overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b \)

\( = |\overrightarrow a|^2 + 2(\overrightarrow a \cdot \overrightarrow b) + |\overrightarrow b|^2 \)

Right side (Magnitude of vector difference):

\( (\overrightarrow a - \overrightarrow b) \cdot (\overrightarrow a - \overrightarrow b) = \overrightarrow a \cdot \overrightarrow a - \overrightarrow a \cdot \overrightarrow b - \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b \)

\( = |\overrightarrow a|^2 - 2(\overrightarrow a \cdot \overrightarrow b) + |\overrightarrow b|^2 \)

Now, we set the expanded expressions equal to each other, as per the given condition \( \left| {\overrightarrow a + \overrightarrow b } \right| = \left| {\overrightarrow a - \overrightarrow b } \right| \):

\( |\overrightarrow a|^2 + 2(\overrightarrow a \cdot \overrightarrow b) + |\overrightarrow b|^2 = |\overrightarrow a|^2 - 2(\overrightarrow a \cdot \overrightarrow b) + |\overrightarrow b|^2 \)

We can subtract \(|\overrightarrow a|^2\) and \(|\overrightarrow b|^2\) from both sides of the equation:

\( 2(\overrightarrow a \cdot \overrightarrow b) = -2(\overrightarrow a \cdot \overrightarrow b) \)

Now, bring the term from the right side to the left side:

\( 2(\overrightarrow a \cdot \overrightarrow b) + 2(\overrightarrow a \cdot \overrightarrow b) = 0 \)

\( 4(\overrightarrow a \cdot \overrightarrow b) = 0 \)

Dividing by 4, we get:

\( \overrightarrow a \cdot \overrightarrow b = 0 \)

This result is key to finding the angle between vectors. Recall the definition of the dot product of two vectors in terms of their magnitudes and the angle between them:

\( \overrightarrow a \cdot \overrightarrow b = |\overrightarrow a| |\overrightarrow b| \cos \theta \)

where \(\theta\) is the angle between \(\overrightarrow a\) and \(\overrightarrow b\). Substituting the result from our previous step:

\( |\overrightarrow a| |\overrightarrow b| \cos \theta = 0 \)

For the product of magnitudes and cosine to be zero, assuming that the vectors \(\overrightarrow a\) and \(\overrightarrow b\) are non-zero vectors (as zero vectors would make the condition trivially true and the angle undefined in some contexts), at least one of the factors must be zero. Since we assume \(|\overrightarrow a| \neq 0\) and \(|\overrightarrow b| \neq 0\), it must be the case that \(\cos \theta = 0\). This condition implies that the angle between vectors \(\overrightarrow a\) and \(\overrightarrow b\) is such that its cosine is zero.

The angles \(\theta\) in the range \(0 \le \theta \le \pi\) for which \(\cos \theta = 0\) are \(\theta = \frac{\pi}{2}\).

Thus, the angle between vectors \(\overrightarrow a\) and \(\overrightarrow b\) must be \( \frac{\pi}{2} \) (or 90 degrees). This means the vectors are orthogonal or perpendicular to each other when the magnitude of their vector sum equals the magnitude of their vector difference.

Let's summarize the steps to find the angle between vectors:

1. Start with the given condition \(|\overrightarrow a + \overrightarrow b| = |\overrightarrow a - \overrightarrow b|\).
2. Square both sides: \(|\overrightarrow a + \overrightarrow b|^2 = |\overrightarrow a - \overrightarrow b|^2\).
3. Expand using the dot product: \((\overrightarrow a + \overrightarrow b) \cdot (\overrightarrow a + \overrightarrow b) = (\overrightarrow a - \overrightarrow b) \cdot (\overrightarrow a - \overrightarrow b)\).
4. Simplify using dot product properties: \(|\overrightarrow a|^2 + 2(\overrightarrow a \cdot \overrightarrow b) + |\overrightarrow b|^2 = |\overrightarrow a|^2 - 2(\overrightarrow a \cdot \overrightarrow b) + |\overrightarrow b|^2\).
5. Cancel terms and rearrange: \(4(\overrightarrow a \cdot \overrightarrow b) = 0\), leading to \(\overrightarrow a \cdot \overrightarrow b = 0\).
6. Use the dot product definition \(\overrightarrow a \cdot \overrightarrow b = |\overrightarrow a| |\overrightarrow b| \cos \theta\) to find the angle.
7. Since \(\overrightarrow a \cdot \overrightarrow b = 0\), we get \(|\overrightarrow a| |\overrightarrow b| \cos \theta = 0\).
8. For non-zero vectors, this implies \(\cos \theta = 0\).
9. The angle between vectors is therefore \( \theta = \frac{\pi}{2} \).

This calculation shows that the angle between vectors satisfying the given condition is always \(\frac{\pi}{2}\).