The vector `cosalpha.cosbetahati+cosalpha.sinbetahatj+sinalphahatk` is a/an

Options

A. Null vector
B. Unit vector
C. Constant vector
D. Vector of magnitude 3

Answer

To determine the nature of the vector given by the expression \( \cos \alpha \cos \beta \hat{i} + \cos \alpha \sin \beta \hat{j} + \sin \alpha \hat{k} \), we will calculate its magnitude and analyze the result.

### Step-by-Step Solution:

1. **Write the vector in standard form**:
\[ \mathbf{v} = \cos \alpha \cos \beta \hat{i} + \cos \alpha \sin \beta \hat{j} + \sin \alpha \hat{k} \]

2. **Identify the coefficients**:
- Coefficient of \( \hat{i} \) is \( a = \cos \alpha \cos \beta \)
- Coefficient of \( \hat{j} \) is \( b = \cos \alpha \sin \beta \)
- Coefficient of \( \hat{k} \) is \( c = \sin \alpha \)

3. **Use the formula for the magnitude of a vector**:
The magnitude \( |\mathbf{v}| \) of a vector \( \mathbf{v} = a \hat{i} + b \hat{j} + c \hat{k} \) is given by:
\[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]

4. **Substitute the coefficients into the formula**:
\[ |\mathbf{v}| = \sqrt{(\cos \alpha \cos \beta)^2 + (\cos \alpha \sin \beta)^2 + (\sin \alpha)^2} \]

5. **Simplify the expression**:
- Calculate \( a^2 + b^2 \):
\[ a^2 + b^2 = \cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta \]
- Factor out \( \cos^2 \alpha \):
\[ = \cos^2 \alpha (\cos^2 \beta + \sin^2 \beta) \]
- Using the Pythagorean identity \( \cos^2 \beta + \sin^2 \beta = 1 \):
\[ = \cos^2 \alpha \cdot 1 = \cos^2 \alpha \]

6. **Combine with \( c^2 \)**:
\[ |\mathbf{v}| = \sqrt{\cos^2 \alpha + \sin^2 \alpha} \]
- Again, using the Pythagorean identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ |\mathbf{v}| = \sqrt{1} = 1 \]

7. **Conclusion**:
Since the magnitude of the vector is 1, we conclude that the vector is a unit vector.

### Final Answer:
The vector \( \cos \alpha \cos \beta \hat{i} + \cos \alpha \sin \beta \hat{j} + \sin \alpha \hat{k} \) is a **unit vector**.