Q.Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes 1 and 2 respectively and when $\vec{a}\cdot\vec{b}=1$.

Concept: Dot Product Angle — the cosine of the angle between two vectors is given by their dot product divided by the product of their magnitudes.

We have $|\vec{a}| = 1$, $|\vec{b}| = 2$, and $\vec{a}\cdot\vec{b} = 1$.

The formula is:

Since $\cos\theta = \frac{1}{2}$, the angle is $\theta = 60^\circ$ (or $\frac{\pi}{3}$ radians).

The dot product of two vectors isn't just a mechanical calculation — it carries geometric meaning. When you take $\vec{a} \cdot \vec{b}$, you're essentially measuring how much one vector "projects" onto the other. The formula $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$ ties this projection to the angle $\theta$ between them. So if you know the magnitudes and the dot product, you can solve for $\cos \theta$, and from there, the angle itself.

Here, we're given $|\vec{a}| = 1$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = 1$. The question is straightforward: find $\theta$.

1. Write the dot product formula The fundamental relation is:

This holds for any two vectors in any dimension — it's the definition of the angle between them.

2. Substitute the known values Plug in $|\vec{a}| = 1$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = 1$:

So:

3. Solve for $\cos \theta$ Divide both sides by 2:

4. Find $\theta$ from the cosine The angle whose cosine is $\frac{1}{2}$ is $60^\circ$ (or $\frac{\pi}{3}$ radians). Since the angle between vectors is conventionally taken between $0^\circ$ and $180^\circ$, this is the unique answer.

Method: Angle between two vectors from the dot product

Use this whenever the "angle between" two vectors is asked, given either components or magnitudes-and-dot-product.

Steps

Step 1: Write the master formula.

You must divide by both magnitudes — $\vec{a}\cdot\vec{b}$ alone equals $\cos\theta$ only for unit vectors.

Step 2: Assemble the three ingredients.

Obtain $\vec{a}\cdot\vec{b}$, $|\vec{a}|$ and $|\vec{b}|$. If components are given, $\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3$ and $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$; here the magnitudes and dot product may be supplied directly.

Step 3: Solve for $\theta$.

Substitute, simplify $\cos\theta$, and take $\theta=\cos^{-1}(\cdot)$. The sign of the result flags the angle type: positive $\Rightarrow$ acute, zero $\Rightarrow$ right angle, negative $\Rightarrow$ obtuse.

Common Mistakes

Mistake 1: Forgetting to divide by both magnitudes.

Why it's wrong: $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$; taking $\vec{a}\cdot\vec{b}$ as $\cos\theta$ is valid only for unit vectors, and here $|\vec{b}|=2$. Correct approach: divide $\vec{a}\cdot\vec{b}=1$ by $|\vec{a}||\vec{b}|=2$, giving $\cos\theta=\frac12$.

Mistake 2: Reading $\cos\theta=\frac12$ as $\theta=30^\circ$.

Why it's wrong: $\cos30^\circ=\frac{\sqrt3}{2}$; the angle with cosine $\frac12$ is $60^\circ$. Correct approach: match the cosine value to the correct standard angle, $\theta=60^\circ=\frac{\pi}{3}$.