Q.Find $|\vec{a}-\vec{b}|$, if two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{a}\cdot\vec{b}=4$.

Concept: Vector Magnitude Difference — use the identity $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\,\vec{a}\cdot\vec{b}$.

Step 1: Write the square of the required magnitude:

Step 2: Substitute the given values $|\vec{a}| = 2$, $|\vec{b}| = 3$, and $\vec{a}\cdot\vec{b} = 4$: …

The key insight here is that the magnitude of a vector difference is not simply the difference of the magnitudes. Instead, it depends on the angle between the vectors — which is captured by the dot product. Think of two sides of a triangle: the vector $\vec{a}-\vec{b}$ is the third side, and the dot product tells you how “spread apart” the two original vectors are.

We are given:

• $|\vec{a}| = 2$
• $|\vec{b}| = 3$
• $\vec{a} \cdot \vec{b} = 4$

We want $|\vec{a} - \vec{b}|$.

1. Start with the square of the magnitude. For any vector, $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. So:

2. Expand the dot product. Using distributivity:

Since dot product is commutative ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$), this simplifies to:

3. Plug in the given numbers.

4. Take the square root. …

Method: Magnitude of a vector difference via the dot-product expansion

Use this to find $|\vec{a}-\vec{b}|$ from the magnitudes and the dot product $\vec{a}\cdot\vec{b}$ (the vector form of the law of cosines).

Steps

Step 1: Square the magnitude and expand.

Since $|\vec{v}|^2=\vec{v}\cdot\vec{v}$,

The cross terms appear twice, giving the factor $2$.

Step 2: Substitute the given data.

Plug in $|\vec{a}|$, $|\vec{b}|$ and $\vec{a}\cdot\vec{b}$ to get a single number for $|\vec{a}-\vec{b}|^2$. …

Common Mistakes

Mistake 1: Writing $|\vec{a}-\vec{b}|=|\vec{a}|-|\vec{b}|=2-3=-1$.

Why it's wrong: the magnitude of a difference is not the difference of magnitudes (and can never be negative); they agree only when the vectors are parallel and same-sense. Correct approach: use $|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}$.

Mistake 2: Forgetting the factor of $2$ on the dot-product term. …