The angles which the vector `A=3hati + 6hatj+2hatk` makes with the co-ordinate axes are

Options

A. `cos^(-1)""(3)/(7)cos^(-1)""(6)/(7)and cos^(-1)""(2)/(7)`
B. `cos^(-1)""(4)/(7)cos^(-1)""(5)/(7)and cos^(-1)""(3)/(7)`
C. `cos^(-1)""(3)/(7)cos^(-1)""(4)/(7)and cos^(-1)""(1)/(7)`
D. None of the above

Explanation

To find the angles that the vector \( \mathbf{A} = 3\hat{i} + 6\hat{j} + 2\hat{k} \) makes with the coordinate axes, we can follow these steps:

### Step 1: Identify the components of the vector
The vector \( \mathbf{A} \) has the following components:
- \( A_x = 3 \) (along the x-axis)
- \( A_y = 6 \) (along the y-axis)
- \( A_z = 2 \) (along the z-axis)

### Step 2: Calculate the magnitude of the vector
The magnitude of vector \( \mathbf{A} \) is calculated using the formula:
\[ |\mathbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Substituting the values:
\[ |\mathbf{A}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 3: Calculate the angles with the coordinate axes
The angles \( \alpha \), \( \beta \), and \( \gamma \) that vector \( \mathbf{A} \) makes with the x-axis, y-axis, and z-axis respectively can be found using the cosine of the angles:
- \( \cos \alpha = \frac{A_x}{|\mathbf{A}|} \)
- \( \cos \beta = \frac{A_y}{|\mathbf{A}|} \)
- \( \cos \gamma = \frac{A_z}{|\mathbf{A}|} \)

Substituting the values:
1. For \( \alpha \):
\[ \cos \alpha = \frac{3}{7} \]
Therefore,
\[ \alpha = \cos^{-1}\left(\frac{3}{7}\right) \]

2. For \( \beta \):
\[ \cos \beta = \frac{6}{7} \]
Therefore,
\[ \beta = \cos^{-1}\left(\frac{6}{7}\right) \]

3. For \( \gamma \):
\[ \cos \gamma = \frac{2}{7} \]
Therefore,
\[ \gamma = \cos^{-1}\left(\frac{2}{7}\right) \]

### Final Result
The angles that the vector \( \mathbf{A} \) makes with the coordinate axes are:
- \( \alpha = \cos^{-1}\left(\frac{3}{7}\right) \)
- \( \beta = \cos^{-1}\left(\frac{6}{7}\right) \)
- \( \gamma = \cos^{-1}\left(\frac{2}{7}\right) \)