Find the angle between two vectors `vec(a)` and `vec(b)` such that :
`|vec(a)|=sqrt(3), |vec(b)|=2` and `vec(a).vec(b)=sqrt(6)`.

Answer

To find the angle \( \theta \) between the two vectors \( \vec{a} \) and \( \vec{b} \), we can use the formula for the dot product of two vectors:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

### Step 1: Substitute the known values into the dot product formula.

We know:
- \( |\vec{a}| = \sqrt{3} \)
- \( |\vec{b}| = 2 \)
- \( \vec{a} \cdot \vec{b} = \sqrt{6} \)

Substituting these values into the dot product formula gives us:

\[ \sqrt{6} = (\sqrt{3})(2) \cos \theta \]

### Step 2: Simplify the equation.

Calculating the right-hand side:

\[ \sqrt{6} = 2\sqrt{3} \cos \theta \]

### Step 3: Isolate \( \cos \theta \).

To isolate \( \cos \theta \), divide both sides by \( 2\sqrt{3} \):

\[ \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} \]

### Step 4: Simplify \( \frac{\sqrt{6}}{2\sqrt{3}} \).

We can simplify this expression:

\[ \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{12}} = \frac{\sqrt{6}}{\sqrt{4 \cdot 3}} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 5: Find \( \theta \).

Now, we can find \( \theta \) by taking the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \]

### Step 6: Calculate the angle.

The angle whose cosine is \( \frac{\sqrt{2}}{2} \) is:

\[ \theta = 45^\circ \text{ or } \frac{\pi}{4} \text{ radians} \]

### Final Answer:

The angle between the two vectors \( \vec{a} \) and \( \vec{b} \) is \( 45^\circ \) or \( \frac{\pi}{4} \) radians.

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