The angle `theta` between the vector `p=hati+hatj +hatk` and unit vector along X-axis is

Options

A. `cos^(-1)((1)/(sqrt(3)))`
B. `cos^(-1)((1)/(sqrt(2)))`
C. `cos^(-1)(sqrt(3)/(2))`
D. `cos^(-1)((1)/(2))`

Explanation

To find the angle \( \theta \) between the vector \( \mathbf{p} = \hat{i} + \hat{j} + \hat{k} \) and the unit vector along the X-axis \( \hat{r} = \hat{i} \), we can use the dot product formula. Here are the steps to solve the problem:

### Step 1: Define the vectors
We have:
- \( \mathbf{p} = \hat{i} + \hat{j} + \hat{k} \)
- \( \hat{r} = \hat{i} \)

### Step 2: Calculate the dot product
The dot product \( \mathbf{p} \cdot \hat{r} \) is calculated as follows:
\[ \mathbf{p} \cdot \hat{r} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} \]
Using the properties of dot products:
\[ \hat{i} \cdot \hat{i} = 1, \quad \hat{j} \cdot \hat{i} = 0, \quad \hat{k} \cdot \hat{i} = 0 \]
Thus,
\[ \mathbf{p} \cdot \hat{r} = 1 + 0 + 0 = 1 \]

### Step 3: Calculate the magnitudes of the vectors
Next, we calculate the magnitudes of \( \mathbf{p} \) and \( \hat{r} \):
- Magnitude of \( \mathbf{p} \):
\[ |\mathbf{p}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
- Magnitude of \( \hat{r} \):
\[ |\hat{r}| = 1 \]

### Step 4: Use the dot product formula
The dot product can also be expressed in terms of the magnitudes and the cosine of the angle \( \theta \):
\[ \mathbf{p} \cdot \hat{r} = |\mathbf{p}| |\hat{r}| \cos \theta \]
Substituting the values we have:
\[ 1 = \sqrt{3} \cdot 1 \cdot \cos \theta \]
This simplifies to:
\[ \cos \theta = \frac{1}{\sqrt{3}} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \]

### Final Answer
Thus, the angle \( \theta \) between the vector \( \mathbf{p} \) and the unit vector along the X-axis is:
\[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \]

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