A particle lies in space at point ( 2,3,4) find the magnitude of its position vector.

Answer

To find the magnitude of the position vector of a particle located at the point (2, 3, 4), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position Vector**:
The position vector \(\vec{OA}\) of a point \(A(x, y, z)\) in space with respect to the origin \(O(0, 0, 0)\) can be expressed as:
\[ \vec{OA} = x \hat{i} + y \hat{j} + z \hat{k} \]
For the point (2, 3, 4), we have:
\[ \vec{OA} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \]

2. **Calculate the Magnitude of the Position Vector**:
The magnitude of a vector \(\vec{A} = a \hat{i} + b \hat{j} + c \hat{k}\) is given by the formula:
\[ |\vec{A}| = \sqrt{a^2 + b^2 + c^2} \]
Here, \(a = 2\), \(b = 3\), and \(c = 4\). Therefore, we calculate:
\[ |\vec{OA}| = \sqrt{2^2 + 3^2 + 4^2} \]

3. **Perform the Squaring**:
Calculate each square:
\[ 2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16 \]

4. **Sum the Squares**:
Now, add these squared values:
\[ 4 + 9 + 16 = 29 \]

5. **Take the Square Root**:
Finally, take the square root of the sum:
\[ |\vec{OA}| = \sqrt{29} \]

6. **Approximate the Value**:
The approximate value of \(\sqrt{29}\) is about:
\[ |\vec{OA}| \approx 5.38 \]

### Final Answer:
The magnitude of the position vector is \(\sqrt{29}\) or approximately \(5.38\).