Q.Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$, respectively having $\vec{a} \cdot \vec{b}=\sqrt{6}$.

Concept: Dot Product Angle — the cosine of the angle between two vectors is given by $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$.

Step 1: Write the formula for the angle between vectors:

Step 2: Substitute the given values:

Step 3: Simplify:

Step 4: Recognise that $\cos\theta = \frac{\sqrt{2}}{2}$ corresponds to $\theta = 45^\circ$ (or $\frac{\pi}{4}$ radians).

The dot product is the bridge between the algebraic components of vectors and their geometric relationship. When you multiply two vectors using the dot product, the result isn't just a number — it encodes how much one vector "projects" onto the other. That projection depends directly on the cosine of the angle between them.

This is why the formula $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$ is so powerful. If you know the magnitudes and the dot product, you can isolate $\cos \theta$ and then find $\theta$ itself. No need to know the components of the vectors at all.

Let's work through it.

1. Write down the dot product formula. For any two vectors $\vec{a}$ and $\vec{b}$, the dot product is:

where $\theta$ is the angle between them.

2. Plug in the given values. We are told $|\vec{a}| = \sqrt{3}$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = \sqrt{6}$. Substituting:

3. Simplify the right-hand side. Multiply the magnitudes:

4. Solve for $\cos \theta$. Divide both sides by $2\sqrt{3}$:

Now simplify the fraction. Notice $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2}\sqrt{3}$. So:

And $\frac{\sqrt{2}}{2}$ is exactly $\frac{1}{\sqrt{2}}$.

5. Find the angle. From trigonometry, $\cos \theta = \frac{1}{\sqrt{2}}$ means $\theta = 45^\circ$ (or $\frac{\pi}{4}$ radians). Since the angle between two vectors is conventionally taken between $0^\circ$ and $180^\circ$, this is the unique answer.

Method: Angle Between Vectors from Magnitudes and Dot Product

Use this when the magnitudes and the dot product are given directly (no components) and you need the angle.

Steps

Step 1: Write the dot-product angle formula.

This isolates the angle from the two "size" pieces and the alignment piece.

Step 2: Substitute the given quantities.

Plug $\vec{a}\cdot\vec{b}$, $|\vec{a}|$ and $|\vec{b}|$ straight in — divide by the product of both magnitudes.

Step 3: Simplify the surd, then recognise the standard cosine.

Reduce the fraction (e.g. $\tfrac{\sqrt6}{2\sqrt3}=\tfrac{1}{\sqrt2}$). Match it to a known value: $\tfrac{1}{\sqrt2}\Rightarrow45^\circ$, $\tfrac12\Rightarrow60^\circ$, and so on.

Step 4: Write the angle, not the cosine.

Give $\theta$ in degrees or radians, taken in $[0^\circ,180^\circ]$.

Common Mistakes

Mistake 1: Dividing by only one magnitude.

Why it's wrong: the denominator is the product $|\vec{a}|\,|\vec{b}|$; omitting one magnitude scales $\cos\theta$ wrongly. Correct approach: divide $\vec{a}\cdot\vec{b}$ by both $|\vec a|$ and $|\vec b|$.

Mistake 2: Stopping at $\cos\theta$ and reporting it as the angle.

Why it's wrong: $\tfrac{1}{\sqrt2}$ is the cosine, not the angle. Correct approach: apply $\cos^{-1}$ to get $\theta=45^\circ$.

Mistake 3: Mis-simplifying $\tfrac{\sqrt6}{2\sqrt3}$.

Why it's wrong: $\tfrac{\sqrt6}{2\sqrt3}=\tfrac{\sqrt2}{2}=\tfrac{1}{\sqrt2}$; leaving it unsimplified hides the standard $45^\circ$ value. Correct approach: rationalise using $\sqrt6=\sqrt2\cdot\sqrt3$.