The displacement of a particle from a point having position vector ` 2hati + 4hatj` to another point having position vector ` 5hatj + 1hatj` is

Options

A. 3 units
B. `3sqrt2` units
C. 5 units
D. ` 5sqrt3`units

Answer

To find the displacement of a particle from one point to another, we need to follow these steps:

### Step 1: Identify the position vectors
The initial position vector \( \mathbf{r_1} \) is given as:
\[ \mathbf{r_1} = 2\hat{i} + 4\hat{j} \]
The final position vector \( \mathbf{r_2} \) is given as:
\[ \mathbf{r_2} = 5\hat{i} + 1\hat{j} \]

### Step 2: Calculate the displacement vector
The displacement vector \( \mathbf{d} \) is calculated by subtracting the initial position vector from the final position vector:
\[ \mathbf{d} = \mathbf{r_2} - \mathbf{r_1} \]

Substituting the values:
\[ \mathbf{d} = (5\hat{i} + 1\hat{j}) - (2\hat{i} + 4\hat{j}) \]

### Step 3: Perform the subtraction
Now, we perform the subtraction component-wise:
\[ \mathbf{d} = (5 - 2)\hat{i} + (1 - 4)\hat{j} \]
\[ \mathbf{d} = 3\hat{i} - 3\hat{j} \]

### Step 4: Calculate the magnitude of the displacement
To find the magnitude of the displacement vector \( \mathbf{d} \):
\[ |\mathbf{d}| = \sqrt{(3)^2 + (-3)^2} \]
\[ |\mathbf{d}| = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Final Answer
The displacement of the particle is:
\[ 3\sqrt{2} \text{ units} \]

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