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

Answer

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

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

where:
- \(|\vec{a}|\) is the magnitude of vector \(\vec{a}\),
- \(|\vec{b}|\) is the magnitude of vector \(\vec{b}\),
- \(\theta\) is the angle between the two vectors.

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

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

Substituting these values into the dot product formula gives:

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

### Step 2: Simplify the equation.

Calculating the right-hand side:

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

### Step 3: Solve for \(\cos \theta\).

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

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

### Step 4: Simplify \(\cos \theta\).

We can simplify \(\frac{\sqrt{6}}{2\sqrt{2}}\):

\[ \cos \theta = \frac{\sqrt{6}}{\sqrt{8}} = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2} \]

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

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

\[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

### Step 6: Determine the angle.

From trigonometric values, we know that:

\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ \]

Thus, the angle between the vectors \(\vec{a}\) and \(\vec{b}\) is:

\[ \theta = 30^\circ \]

### Final Answer:
The angle between \(\vec{a}\) and \(\vec{b}\) is \(30^\circ\).

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