The angle between the vectors `2i+3j+k and 2i-j-k` is

Options

A. `pi//2`
B. `pi//4`
C. `pi//3`
D. 0

Answer

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

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

### Step 1: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)

The dot product is calculated as follows:

\[ \mathbf{A} \cdot \mathbf{B} = (2)(2) + (3)(-1) + (1)(-1) \]

Calculating each term:

- \( 2 \times 2 = 4 \)
- \( 3 \times -1 = -3 \)
- \( 1 \times -1 = -1 \)

Adding these results together:

\[ \mathbf{A} \cdot \mathbf{B} = 4 - 3 - 1 = 0 \]

### Step 2: Calculate the magnitudes \( |\mathbf{A}| \) and \( |\mathbf{B}| \)

The magnitude of vector \( \mathbf{A} \) is given by:

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

The magnitude of vector \( \mathbf{B} \) is given by:

\[ |\mathbf{B}| = \sqrt{(2)^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 3: Substitute the values into the cosine formula

Now, we can substitute the values into the cosine formula:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{0}{\sqrt{14} \cdot \sqrt{6}} = 0 \]

### Step 4: Determine the angle \( \theta \)

Since \( \cos \theta = 0 \), we know that:

\[ \theta = 90^\circ \]

### Final Answer

The angle between the vectors \( 2\mathbf{i} + 3\mathbf{j} + \mathbf{k} \) and \( 2\mathbf{i} - \mathbf{j} - \mathbf{k} \) is \( 90^\circ \).

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