Q.Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

Expand $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$ as a difference of squares — the cross terms cancel.

Step 1 — Expand. $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = 8.$

Step 2 — Use $|\vec{a}|=8|\vec{b}|$. Then $|\vec{a}|^2 = 64|\vec{b}|^2$, so $64|\vec{b}|^2 - |\vec{b}|^2 = 63|\vec{b}|^2 = 8$, giving $|\vec{b}|^2 = \dfrac{8}{63}$. …

The idea

A dot product of a sum and a difference behaves just like the algebraic identity $(x+y)(x-y)=x^2-y^2$. For vectors,

Because the dot product is commutative, $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$, so the two middle terms cancel and only the squared magnitudes survive.

Set up the equations

1. Expand the given product.

2. Bring in the magnitude relation. We are told $|\vec{a}| = 8|\vec{b}|$, so $|\vec{a}|^2 = 64|\vec{b}|^2$. Substituting,

3. Solve for $|\vec{b}|$. Since a magnitude is non-negative, …

Method: Using the Difference-of-Squares Dot-Product Identity

Use this whenever a product like $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$ appears together with magnitude conditions.

Steps

Step 1: Expand using the algebraic identity.

Because the dot product is commutative, the cross terms cancel exactly like ordinary algebra:

Step 2: Set the expansion equal to the given value.

Here $|\vec{a}|^2-|\vec{b}|^2=$ (given number). This is one equation in two unknowns.

Step 3: Bring in the second condition to eliminate one unknown. …

Common Mistakes

Mistake 1: Keeping a cross term $-2\vec{a}\cdot\vec{b}$.

Why it's wrong: $(\vec a+\vec b)\cdot(\vec a-\vec b)$ is a difference of squares, so the $\vec a\cdot\vec b$ terms cancel — it equals $|\vec a|^2-|\vec b|^2$, not $|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2$ (that is $|\vec a-\vec b|^2$). Correct approach: expand as $|\vec a|^2-|\vec b|^2$.

Mistake 2: Forgetting to square the relation $|\vec a|=8|\vec b|$. …