To prove that vectors [tex]\(\vec{x}\)[/tex] and [tex]\(\vec{y}\)[/tex] are perpendicular to each other given the equation [tex]\((\vec{x} + \vec{y})^2 = (\vec{x} - \vec{y})^2\)[/tex], we will follow these steps:
1. Expand both sides of the equation.
2. Simplify the equation to find a relationship between [tex]\(\vec{x}\)[/tex] and [tex]\(\vec{y}\)[/tex].
Step 1: Expand both sides
The given equation is:
[tex]\[
(\vec{x} + \vec{y})^2 = (\vec{x} - \vec{y})^2
\][/tex]
First, we need to expand each side separately using the distributive property and the properties of dot products.
The left-hand side expansion:
[tex]\[
(\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = \vec{x} \cdot \vec{x} + 2 \vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{y}
\][/tex]
Simplifying this, we get:
[tex]\[
\|\vec{x}\|^2 + 2 (\vec{x} \cdot \vec{y}) + \|\vec{y}\|^2
\][/tex]
The right-hand side expansion:
[tex]\[
(\vec{x} - \vec{y}) \cdot (\vec{x} - \vec{y}) = \vec{x} \cdot \vec{x} - 2 \vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{y}
\][/tex]
Simplifying this, we get:
[tex]\[
\|\vec{x}\|^2 - 2 (\vec{x} \cdot \vec{y}) + \|\vec{y}\|^2
\][/tex]
Step 2: Set the expanded forms equal and simplify
Now we set the expanded forms of both sides equal to each other:
[tex]\[
\|\vec{x}\|^2 + 2 (\vec{x} \cdot \vec{y}) + \|\vec{y}\|^2 = \|\vec{x}\|^2 - 2 (\vec{x} \cdot \vec{y}) + \|\vec{y}\|^2
\][/tex]
Subtract [tex]\(\|\vec{x}\|^2 + \|\vec{y}\|^2\)[/tex] from both sides of the equation:
[tex]\[
2 (\vec{x} \cdot \vec{y}) = -2 (\vec{x} \cdot \vec{y})
\][/tex]
Combine like terms:
[tex]\[
4 (\vec{x} \cdot \vec{y}) = 0
\][/tex]
Dividing both sides by 4, we get:
[tex]\[
\vec{x} \cdot \vec{y} = 0
\][/tex]
Conclusion:
The dot product [tex]\(\vec{x} \cdot \vec{y} = 0\)[/tex] implies that the vectors [tex]\(\vec{x}\)[/tex] and [tex]\(\vec{y}\)[/tex] are perpendicular to each other. Thus, we have proven that [tex]\(\vec{x}\)[/tex] and [tex]\(\vec{y}\)[/tex] are perpendicular to each other.
