Q.Find the magnitude of each of two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude such that the angle between them is $60^{\circ}$ and their scalar product is $\dfrac{9}{2}$.

Finding the Angle Between Vectors

Suppose you have two arrows drawn from the same point. One question is unavoidable in geometry, physics, and mechanics: what is the angle between them? You could measure it with a protractor on paper, but that fails the moment the vectors live in 3D. The dot product gives you the angle by pure calculation.

The Core Idea

The scalar (dot) product of two vectors has two faces that describe the same number:

The first is easy to compute from coordinates; the second hides the angle $\theta$ (with $0 \le \theta \le \pi$) between the vectors. Setting them equal and solving for $\cos\theta$ gives the master formula.

Why It Works

Both vectors have a fixed length, so the only thing the dot product can "vary" with is how aligned they are. When they point the same way, $\cos\theta = 1$ and the dot product is as large as possible, $|\vec{a}||\vec{b}|$. When they are perpendicular, $\cos\theta = 0$ and the dot product vanishes. When they point opposite ways, $\cos\theta = -1$. Dividing $\vec{a}\cdot\vec{b}$ by the two lengths simply strips away the size information and leaves behind a pure measure of alignment — exactly $\cos\theta$.

Using the Formula

For $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 0\hat{j} + 0\hat{k}$:

$\cos\theta = \frac{1}{3\cdot 1} = \frac{1}{3} \;\Rightarrow\; \theta = \cos^{-1}\tfrac{1}{3} \approx 70.5^\circ$ …