Vector `P=6hati+4sqrt(2)hatj+4sqrt(2)hatk` makes angle from Z-axas equal to

Options

A. `cos^(-1)(sqrt(2)/(5))`
B. `cos^(-1)(2sqrt(2))`
C. `cos^(-1)""(2sqrt(2))/(5)`
D. None of these

Explanation

To find the angle that vector \( \mathbf{P} = 6\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k} \) makes with the Z-axis, we can follow these steps:

### Step 1: Identify the components of the vector
The vector \( \mathbf{P} \) has the following components:
- \( P_x = 6 \) (i-component)
- \( P_y = 4\sqrt{2} \) (j-component)
- \( P_z = 4\sqrt{2} \) (k-component)

### Step 2: Calculate the magnitude of the vector
The magnitude of vector \( \mathbf{P} \) is given by the formula:
\[ |\mathbf{P}| = \sqrt{P_x^2 + P_y^2 + P_z^2} \]
Substituting the values:
\[ |\mathbf{P}| = \sqrt{6^2 + (4\sqrt{2})^2 + (4\sqrt{2})^2} \]
Calculating each term:
- \( 6^2 = 36 \)
- \( (4\sqrt{2})^2 = 16 \times 2 = 32 \)

Thus,
\[ |\mathbf{P}| = \sqrt{36 + 32 + 32} = \sqrt{100} = 10 \]

### Step 3: Find the cosine of the angle with the Z-axis
The cosine of the angle \( \gamma \) that the vector makes with the Z-axis can be calculated using the formula:
\[ \cos \gamma = \frac{P_z}{|\mathbf{P}|} \]
Substituting the values we have:
\[ \cos \gamma = \frac{4\sqrt{2}}{10} = \frac{2\sqrt{2}}{5} \]

### Step 4: Calculate the angle \( \gamma \)
To find the angle \( \gamma \), we take the inverse cosine:
\[ \gamma = \cos^{-1}\left(\frac{2\sqrt{2}}{5}\right) \]

### Final Answer
The angle \( \gamma \) that vector \( \mathbf{P} \) makes with the Z-axis is:
\[ \gamma = \cos^{-1}\left(\frac{2\sqrt{2}}{5}\right) \]