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, we can follow these steps:

### Step 1: Identify the components of the vector
The vector \(\vec{A}\) can be expressed in terms of its components:
\[ \vec{A} = \hat{i} + \hat{j} + \hat{k} \]
This means that the components of the vector are:
- \(A_x = 1\) (component along the X-axis)
- \(A_y = 1\) (component along the Y-axis)
- \(A_z = 1\) (component along the Z-axis)

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

### Step 3: Calculate the direction cosines
The direction cosines are defined as:
\[ \cos \alpha = \frac{A_x}{|\vec{A}|}, \quad \cos \beta = \frac{A_y}{|\vec{A}|}, \quad \cos \gamma = \frac{A_z}{|\vec{A}|} \]
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: Find the angles using the inverse cosine function
Now, we can find the angles \(\alpha\), \(\beta\), and \(\gamma\):
\[ \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right), \quad \beta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right), \quad \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

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