The vector `cos alpha cos beta hati + cos alpha sin beta hatj + sin alpha hatk ` is a

Options

A. null vector
B. unit vector
C. constant vector
D. none of these

Answer

To solve the problem, we need to determine the nature of the given vector:

\[ \mathbf{A} = \cos \alpha \cos \beta \, \hat{i} + \cos \alpha \sin \beta \, \hat{j} + \sin \alpha \, \hat{k} \]

### Step 1: Identify the components of the vector
The vector \(\mathbf{A}\) has three components:
- \(x = \cos \alpha \cos \beta\)
- \(y = \cos \alpha \sin \beta\)
- \(z = \sin \alpha\)

### Step 2: Calculate the magnitude of the vector
The magnitude of a vector \(\mathbf{A} = x \hat{i} + y \hat{j} + z \hat{k}\) is given by the formula:

\[ |\mathbf{A}| = \sqrt{x^2 + y^2 + z^2} \]

Substituting the components into the formula:

\[ |\mathbf{A}| = \sqrt{(\cos \alpha \cos \beta)^2 + (\cos \alpha \sin \beta)^2 + (\sin \alpha)^2} \]

### Step 3: Simplify the expression
Now we simplify the expression inside the square root:

\[ |\mathbf{A}| = \sqrt{\cos^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + \sin^2 \alpha} \]

### Step 4: Factor out common terms
Notice that \(\cos^2 \alpha\) is common in the first two terms:

\[ |\mathbf{A}| = \sqrt{\cos^2 \alpha (\cos^2 \beta + \sin^2 \beta) + \sin^2 \alpha} \]

### Step 5: Use the Pythagorean identity
Using the identity \(\cos^2 \beta + \sin^2 \beta = 1\):

\[ |\mathbf{A}| = \sqrt{\cos^2 \alpha \cdot 1 + \sin^2 \alpha} = \sqrt{\cos^2 \alpha + \sin^2 \alpha} \]

### Step 6: Apply the Pythagorean identity again
Using the identity \(\cos^2 \alpha + \sin^2 \alpha = 1\):

\[ |\mathbf{A}| = \sqrt{1} = 1 \]

### Conclusion
Since the magnitude of the vector \(\mathbf{A}\) is 1, we can conclude that \(\mathbf{A}\) is a unit vector.

Thus, the correct answer is that the vector is a **unit vector**.

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