Q.If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=8$, then find $|\vec{x}|$.

Concept: Vector Magnitude Properties — using $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$ and the fact that for a unit vector $\vec{a}$, $|\vec{a}| = 1$.

Step 1: Expand the dot product:

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

The key here is to treat the dot product like ordinary algebra — but with vectors. The expression $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})$ looks like a difference of squares, and that’s exactly how it behaves, because the dot product is bilinear and commutative.

We also know $\vec{a}$ is a unit vector, so $|\vec{a}| = 1$. That’s the only extra fact we need.

Let’s go step by step.

1. Expand the dot product Using the distributive property:

Since $\vec{x}\cdot\vec{a} = \vec{a}\cdot\vec{x}$, the middle two terms cancel:

2. Rewrite in terms of magnitudes For any vector $\vec{v}$, $\vec{v}\cdot\vec{v} = |\vec{v}|^2$. So:

3. Plug in the given values The problem states this equals $8$, and $\vec{a}$ is a unit vector so $|\vec{a}|=1$:

4. Solve for $|\vec{x}|$ …

Method: Solving a vector equation using the difference-of-squares identity

Use this when a scalar condition of the form $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=k$ must be solved for $|\vec{x}|$.

Steps

Step 1: Expand the dot product like ordinary algebra.

The dot product is distributive and commutative, so

The two middle terms are equal and cancel (they do not double).

Step 2: Replace self-dot-products with squared magnitudes.

Using $\vec{v}\cdot\vec{v}=|\vec{v}|^2$: …

Common Mistakes

Mistake 1: Doubling the cross terms instead of cancelling them.

Why it's wrong: $\vec{x}\cdot\vec{a}$ and $\vec{a}\cdot\vec{x}$ are equal with opposite signs in this expansion, so they cancel — they do not add to $2\vec{x}\cdot\vec{a}$. Correct approach: $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=|\vec{x}|^2-|\vec{a}|^2$. …