Find out the angle made by `vecA=hati+hatj+hatk` vector from X,Y and Z axes respectively.

Explanation

To find the angles made by the vector \(\vec{A} = \hat{i} + \hat{j} + \hat{k}\) with the X, Y, and Z axes respectively, we can follow these steps:

### Step 1: Identify the vector and its components
The vector given is:
\[ \vec{A} = \hat{i} + \hat{j} + \hat{k} \]
This means the components of the vector are:
- \(A_x = 1\) (coefficient of \(\hat{i}\))
- \(A_y = 1\) (coefficient of \(\hat{j}\))
- \(A_z = 1\) (coefficient of \(\hat{k}\))

### Step 2: Calculate the magnitude of the vector
The magnitude of the vector \(\vec{A}\) is calculated using the formula:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Substituting the values:
\[ |\vec{A}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

### Step 3: Find the direction cosines
The direction cosines are given by:
\[ \cos \alpha = \frac{A_x}{|\vec{A}|}, \quad \cos \beta = \frac{A_y}{|\vec{A}|}, \quad \cos \gamma = \frac{A_z}{|\vec{A}|} \]
Substituting the values:
\[ \cos \alpha = \frac{1}{\sqrt{3}}, \quad \cos \beta = \frac{1}{\sqrt{3}}, \quad \cos \gamma = \frac{1}{\sqrt{3}} \]

### Step 4: Calculate the angles with the axes
To find the angles \(\alpha\), \(\beta\), and \(\gamma\), we take the inverse cosine of the direction cosines:
\[ \alpha = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right), \quad \beta = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right), \quad \gamma = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right) \]

### Final Result
Thus, the angles made by the vector \(\vec{A}\) with the X, Y, and Z axes are:
\[ \alpha = \beta = \gamma = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right) \]