The angles with a vector `hat(i)+hat(j)+sqrt(2hat(k))` makes with X,Y and Z axes respectively are

Options

A. `60^@`,`60^@`,`60^@`
B. `45^@`,`45^@`,`45^@`
C. `60^@`,`60^@`,`45^@`
D. `45^@`,`45^@`,`60^@`

Explanation

To find the angles that the vector \(\hat{i} + \hat{j} + \sqrt{2} \hat{k}\) makes with the X, Y, and Z axes, we can follow these steps:

### Step 1: Identify the vector components
The vector can be expressed in component form as:
\[ \vec{A} = \hat{i} + \hat{j} + \sqrt{2} \hat{k} \]
This means:
- \(A_x = 1\) (component along x-axis)
- \(A_y = 1\) (component along y-axis)
- \(A_z = \sqrt{2}\) (component along z-axis)

### Step 2: Calculate the magnitude of the vector
The magnitude \(r\) of the vector \(\vec{A}\) is given by:
\[ r = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Substituting the values:
\[ r = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2 \]

### Step 3: Calculate the direction cosines
The direction cosines are given by:
- For the x-axis:
\[ \cos \alpha = \frac{A_x}{r} = \frac{1}{2} \]
- For the y-axis:
\[ \cos \beta = \frac{A_y}{r} = \frac{1}{2} \]
- For the z-axis:
\[ \cos \gamma = \frac{A_z}{r} = \frac{\sqrt{2}}{2} \]

### Step 4: Find the angles using the inverse cosine function
Now, we can find the angles:
- For the x-axis:
\[ \alpha = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]
- For the y-axis:
\[ \beta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]
- For the z-axis:
\[ \gamma = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45^\circ \]

### Final Result
The angles that the vector makes with the X, Y, and Z axes are:
- \(\alpha = 60^\circ\)
- \(\beta = 60^\circ\)
- \(\gamma = 45^\circ\)

### Conclusion
Thus, the angles with the X, Y, and Z axes respectively are \(60^\circ, 60^\circ, 45^\circ\).

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