A particle moves `3 m` north, then `4 m` east, and then `12 m` vertically upwards, its displacement is

Options

A. `12 m`
B. `13 m`
C. `19 m`
D. `15 m`

Answer

To find the displacement of the particle that moves in three dimensions, we can follow these steps:

### Step 1: Identify the movements
The particle moves:
- 3 meters north (positive y-direction)
- 4 meters east (positive x-direction)
- 12 meters vertically upwards (positive z-direction)

### Step 2: Represent the movements in vector form
We can represent these movements using unit vectors:
- North (y-direction) is represented as \(3 \hat{j}\)
- East (x-direction) is represented as \(4 \hat{i}\)
- Upwards (z-direction) is represented as \(12 \hat{k}\)

Thus, the position vector \( \vec{r} \) of the particle can be expressed as:
\[ \vec{r} = 4 \hat{i} + 3 \hat{j} + 12 \hat{k} \]

### Step 3: Calculate the magnitude of the displacement vector
The magnitude of the displacement vector \( |\vec{r}| \) can be calculated using the formula:
\[ |\vec{r}| = \sqrt{x^2 + y^2 + z^2} \]
where \( x = 4 \), \( y = 3 \), and \( z = 12 \).

Substituting the values:
\[ |\vec{r}| = \sqrt{(4)^2 + (3)^2 + (12)^2} \]
\[ |\vec{r}| = \sqrt{16 + 9 + 144} \]
\[ |\vec{r}| = \sqrt{169} \]
\[ |\vec{r}| = 13 \text{ meters} \]

### Conclusion
The displacement of the particle is \( 13 \text{ meters} \).

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