A Body moves 6m north, 8m east and 10m vertically upwards, what is its resultant displacement from initial position

Options

A. `10sqrt(2)` m
B. 10 m
C. `(10)/(sqrt(2))` m
D. 20 m

Explanation

To solve the problem of finding the resultant displacement of a body that moves 6m north, 8m east, and 10m vertically upwards, we can follow these steps:

### Step 1: Define the Displacement Vectors
The body moves in three dimensions:
- 8m east can be represented as \( \vec{R_1} = 8 \hat{i} \) (where \( \hat{i} \) is the unit vector in the east direction).
- 6m north can be represented as \( \vec{R_2} = 6 \hat{j} \) (where \( \hat{j} \) is the unit vector in the north direction).
- 10m vertically upwards can be represented as \( \vec{R_3} = 10 \hat{k} \) (where \( \hat{k} \) is the unit vector in the upward direction).

### Step 2: Calculate the Resultant Displacement Vector
The resultant displacement vector \( \vec{R} \) can be calculated by adding the individual displacement vectors:
\[ \vec{R} = \vec{R_1} + \vec{R_2} + \vec{R_3} = 8 \hat{i} + 6 \hat{j} + 10 \hat{k} \]

### Step 3: Calculate the Magnitude of the Resultant Displacement
To find the magnitude of the resultant displacement vector \( \vec{R} \), we use the formula for the magnitude of a vector in three dimensions:
\[ |\vec{R}| = \sqrt{(R_x)^2 + (R_y)^2 + (R_z)^2} \]
Substituting the values:
\[ |\vec{R}| = \sqrt{(8)^2 + (6)^2 + (10)^2} = \sqrt{64 + 36 + 100} \]
\[ |\vec{R}| = \sqrt{200} \]

### Step 4: Simplify the Magnitude
We can simplify \( \sqrt{200} \):
\[ \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \]

### Conclusion
The resultant displacement from the initial position is:
\[ 10\sqrt{2} \text{ meters} \]