Q.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 $\vec{b}.$

Concept: Perpendicular Vectors Condition — two vectors are perpendicular iff their dot product is zero.

Let $\vec{u} = |\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ and $\vec{v} = |\vec{a}|\vec{b} - |\vec{b}|\vec{a}$.

Compute $\vec{u} \cdot \vec{v}$:

Expand using distributivity:

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

Why This Approach Works

For any two nonzero vectors $\vec{a}$ and $\vec{b}$, we want to prove that the vectors $\vec{u} = |\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ and $\vec{v} = |\vec{a}|\vec{b} - |\vec{b}|\vec{a}$ are perpendicular. The fundamental condition for perpendicularity (orthogonality) is that their dot product equals zero: $\vec{u} \cdot \vec{v} = 0$.

The trick here is that the coefficients $|\vec{a}|$ and $|\vec{b}|$ are scalars (numbers), so they can be pulled out of dot products freely. The expression will simplify beautifully because the cross-terms cancel — a classic pattern where $(A+B)\cdot(A-B) = A\cdot A - B\cdot B$.

Step-by-Step Solution

1. Define the two vectors clearly.

Let:

We need to show $\vec{u} \cdot \vec{v} = 0$.

2. Compute the dot product $\vec{u} \cdot \vec{v}$.

Using the distributive property of the dot product:

This expands as:

3. Factor out the scalar coefficients.

Remember that for any scalars $p, q$ and vectors $\vec{x}, \vec{y}$, we have $(p\vec{x})\cdot(q\vec{y}) = pq (\vec{x}\cdot\vec{y})$. Applying this:

• First term: $(|\vec{a}|\vec{b})\cdot(|\vec{a}|\vec{b}) = |\vec{a}|^2 (\vec{b}\cdot\vec{b}) = |\vec{a}|^2 |\vec{b}|^2$
• Second term: $(|\vec{a}|\vec{b})\cdot(|\vec{b}|\vec{a}) = |\vec{a}||\vec{b}| (\vec{b}\cdot\vec{a})$
• Third term: $(|\vec{b}|\vec{a})\cdot(|\vec{a}|\vec{b}) = |\vec{b}||\vec{a}| (\vec{a}\cdot\vec{b})$
• Fourth term: $(|\vec{b}|\vec{a})\cdot(|\vec{b}|\vec{a}) = |\vec{b}|^2 (\vec{a}\cdot\vec{a}) = |\vec{b}|^2 |\vec{a}|^2$

So: …

Method: Proving Perpendicularity by a Vanishing Dot Product

To prove two vectors are perpendicular — even ones built symbolically from other vectors — show their dot product is zero; you never need components or an actual angle.

Steps

Step 1: Set up the target dot product.

Name the two vectors $\vec{u}$ and $\vec{v}$ and write out $\vec{u}\cdot\vec{v}$. Perpendicularity is proved once this equals $0$.

Step 2: Expand, pulling scalar coefficients out. …

Common Mistakes

Mistake 1: Treating $|\vec{a}|\vec{b}$ as if $|\vec{a}|$ were a vector.

Why it's wrong: $|\vec{a}|$ is a scalar (a number), so $(|\vec{a}|\vec{b})\cdot(|\vec{b}|\vec{a})=|\vec{a}||\vec{b}|(\vec{b}\cdot\vec{a})$. Correct approach: factor scalars cleanly out of the dot product.

Mistake 2: Confusing $\vec{a}\cdot\vec{a}$ with $|\vec{a}|$. …