Find the angle between the vectors `bari+2barj+3bark" and "3bari-barj+2bark`.

Answer

To find the angle between the vectors \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) and \( \mathbf{b} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k} \), we can use the formula for the dot product of two vectors:

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

where \( \theta \) is the angle between the two vectors.

### Step 1: Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \)

The dot product is calculated as follows:

\[ \mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(-1) + (3)(2) \]

Calculating each term:

\[ = 3 - 2 + 6 = 7 \]

### Step 2: Calculate the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \)

**Magnitude of \( \mathbf{a} \)**:

\[ |\mathbf{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

**Magnitude of \( \mathbf{b} \)**:

\[ |\mathbf{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

### Step 3: Substitute into the dot product formula

Now we substitute the values into the dot product formula:

\[ 7 = |\mathbf{a}| |\mathbf{b}| \cos \theta \]

Substituting the magnitudes:

\[ 7 = \sqrt{14} \cdot \sqrt{14} \cdot \cos \theta \]

This simplifies to:

\[ 7 = 14 \cos \theta \]

### Step 4: Solve for \( \cos \theta \)

Dividing both sides by 14:

\[ \cos \theta = \frac{7}{14} = \frac{1}{2} \]

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

Now we find the angle \( \theta \):

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

The angle whose cosine is \( \frac{1}{2} \) is:

\[ \theta = \frac{\pi}{3} \text{ radians} \quad \text{or} \quad 60^\circ \]

### Final Answer

The angle between the vectors \( \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) and \( 3\mathbf{i} - \mathbf{j} + 2\mathbf{k} \) is \( \frac{\pi}{3} \) radians or \( 60^\circ \).

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