Q11:

Show that $|\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b} - |\vec{b}|\vec{a}$, for any two nonzero vectors $\vec{a}$ and $\vec{b}$.

Step 1: Theoretical Foundation

Two vectors are perpendicular if and only if their scalar (dot) product is equal to zero. Let the two given vectors be defined as follows:

Let $\vec{u} = |\vec{a}|\vec{b} + |\vec{b}|\vec{a}$

Let $\vec{v} = |\vec{a}|\vec{b} - |\vec{b}|\vec{a}$

To prove $\vec{u} \perp \vec{v}$, we must demonstrate that $\vec{u} \cdot \vec{v} = 0$ [By Definition of Orthogonal Vectors].

Step 2: Performing the Scalar Product

We calculate the dot product $\vec{u} \cdot \vec{v}$ using the distributive property of the dot product over vector addition: $\vec{A} \cdot (\vec{B} - \vec{C}) = \vec{A} \cdot \vec{B} - \vec{A} \cdot \vec{C}$.

$\vec{u} \cdot \vec{v} = (|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} - |\vec{b}|\vec{a})$

Expanding the product [By Distributive Law]:

$= (|\vec{a}|\vec{b}) \cdot (|\vec{a}|\vec{b}) - (|\vec{a}|\vec{b}) \cdot (|\vec{b}|\vec{a}) + (|\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b}) - (|\vec{b}|\vec{a}) \cdot (|\vec{b}|\vec{a})$

Step 3: Algebraic Simplification

Utilizing the properties of scalar multiplication with vectors ($k\vec{A} \cdot m\vec{B} = km(\vec{A} \cdot \vec{B})$) and the property $\vec{A} \cdot \vec{A} = |\vec{A}|^2$:

$= |\vec{a}|^2(\vec{b} \cdot \vec{b}) - |\vec{a}||\vec{b}|(\vec{b} \cdot \vec{a}) + |\vec{b}||\vec{a}|(\vec{a} \cdot \vec{b}) - |\vec{b}|^2(\vec{a} \cdot \vec{a})$

Since the dot product is commutative ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$) [By Commutative Property of Dot Product]:

$= |\vec{a}|^2|\vec{b}|^2 - |\vec{a}||\vec{b}|(\vec{a} \cdot \vec{b}) + |\vec{a}||\vec{b}|(\vec{a} \cdot \vec{b}) - |\vec{b}|^2|\vec{a}|^2$

Step 4: Cancellation and Final Verification

Observing the terms:

$= (|\vec{a}|^2|\vec{b}|^2 - |\vec{b}|^2|\vec{a}|^2) + (- |\vec{a}||\vec{b}|(\vec{a} \cdot \vec{b}) + |\vec{a}||\vec{b}|(\vec{a} \cdot \vec{b}))$

$= 0 + 0$

$= 0$

Final Solution: Since the scalar product of $(|\vec{a}|\vec{b} + |\vec{b}|\vec{a})$ and $(|\vec{a}|\vec{b} - |\vec{b}|\vec{a})$ is 0, the two vectors are confirmed to be perpendicular.