Q.Find $|\vec{x}|,$ if for a unit vector $\vec{a}, (\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a})=12.$

Concept: Vector Magnitude Properties — using the dot product identity $(\vec{u} - \vec{v})\cdot(\vec{u} + \vec{v}) = |\vec{u}|^2 - |\vec{v}|^2$.

Step 1: Expand the given dot product:

Step 2: Since $\vec{x}\cdot\vec{a} = \vec{a}\cdot\vec{x}$, the middle terms cancel, leaving: …

We start with the given equation:

$(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$, where $\vec{a}$ is a unit vector.

The core insight here is that the dot product of two vectors is distributive over addition, just like ordinary multiplication. This lets us expand the expression into a sum of simpler dot products. Once expanded, we’ll use the property that for any vector $\vec{v}$, $\vec{v} \cdot \vec{v} = |\vec{v}|^2$. Since $\vec{a}$ is a unit vector, $|\vec{a}| = 1$, so $\vec{a} \cdot \vec{a} = 1$.

Let’s work through it step by step.

1. Expand the dot product

   Using the distributive property:

   $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a}$.

2. Simplify the cross terms

   Notice that $\vec{x} \cdot \vec{a}$ and $\vec{a} \cdot \vec{x}$ are equal (dot product is commutative). So:

   $\vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} = 0$.

   The expression reduces to:

   $\vec{x} \cdot \vec{x} - \vec{a} \cdot \vec{a}$.

3. Replace with magnitudes

   $\vec{x} \cdot \vec{x} = |\vec{x}|^2$ and $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.

   Since $\vec{a}$ is a unit vector, $|\vec{a}| = 1$, so $|\vec{a}|^2 = 1$.

   Thus, the equation becomes:

   $|\vec{x}|^2 - 1 = 12$. …

Method: The Difference-of-Squares Trick for Dot Products

Use this for any expression of the shape $(\vec{u}-\vec{v})\cdot(\vec{u}+\vec{v})$ — it collapses just like the scalar identity $(p-q)(p+q)=p^2-q^2$.

Steps

Step 1: Expand with the distributive law.

Step 2: Cancel the cross terms. …

Common Mistakes

Mistake 1: Mishandling the cross terms.

Why it's wrong: some keep a leftover $2\,\vec{x}\cdot\vec{a}$ term. In $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})$ the terms $+\vec{x}\cdot\vec{a}$ and $-\vec{a}\cdot\vec{x}$ are equal and opposite. Correct approach: they cancel, leaving $|\vec{x}|^2-|\vec{a}|^2$.

Mistake 2: Forgetting $\vec{a}$ is a unit vector. …