The angle btween two vectors `2hati + 3 hatj + hatk and - 3hati + 6 hatk ` is

Options

A. `0^@`
B. `45^@`
C. `60^@`
D. `90^@`

Explanation

To find the angle between the two vectors \( \mathbf{A} = 2\hat{i} + 3\hat{j} + \hat{k} \) and \( \mathbf{B} = -3\hat{i} + 6\hat{k} \), 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 of two vectors \( \mathbf{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \mathbf{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) is given by:

\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

For our vectors:

- \( a_1 = 2, a_2 = 3, a_3 = 1 \)
- \( b_1 = -3, b_2 = 0, b_3 = 6 \)

Calculating the dot product:

\[ \mathbf{A} \cdot \mathbf{B} = (2)(-3) + (3)(0) + (1)(6) = -6 + 0 + 6 = 0 \]

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

The magnitude of a vector \( \mathbf{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) is given by:

\[ |\mathbf{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

Calculating \( |\mathbf{A}| \):

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

Calculating \( |\mathbf{B}| \):

\[ |\mathbf{B}| = \sqrt{(-3)^2 + 0^2 + 6^2} = \sqrt{9 + 0 + 36} = \sqrt{45} = 3\sqrt{5} \]

### Step 3: Substitute into the cosine formula

Now substituting the values into the cosine formula:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{0}{\sqrt{14} \cdot 3\sqrt{5}} = 0 \]

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

Since \( \cos \theta = 0 \), we have:

\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{or} \quad 90^\circ \]

### Final Answer

The angle between the two vectors is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

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