A particle moves `21m` along the vector `6hat(i)+2hat(j)+3hat(k)` , then `14 m` along the vector `3hat(i)-2hat(j)+6hat(k)`. Its total displacement (in meters) is

Options

A. `15hat(i)+2hat(j)+12hat(k)`
B. `9hat(i)+12hat(k)`
C. `9hat(i)+6hat(j)`
D. `24hat(i)+2hat(j)+21hat(k)`

Explanation

To find the total displacement of the particle, we can break down the problem into the following steps:

### Step 1: Calculate the unit vector for the first displacement
The particle moves 21 m along the vector \( \vec{A} = 6\hat{i} + 2\hat{j} + 3\hat{k} \).

First, we need to find the magnitude of this vector:
\[ |\vec{A}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \]

Next, we find the unit vector \( \hat{u}_A \):
\[ \hat{u}_A = \frac{\vec{A}}{|\vec{A}|} = \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} \]

### Step 2: Calculate the displacement vector for the first movement
Now, we calculate the displacement vector \( \vec{AB} \):
\[ \vec{AB} = 21 \cdot \hat{u}_A = 21 \cdot \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = 3 \cdot (6\hat{i} + 2\hat{j} + 3\hat{k}) = 18\hat{i} + 6\hat{j} + 9\hat{k} \]

### Step 3: Calculate the unit vector for the second displacement
The particle then moves 14 m along the vector \( \vec{B} = 3\hat{i} - 2\hat{j} + 6\hat{k} \).

First, we find the magnitude of this vector:
\[ |\vec{B}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]

Next, we find the unit vector \( \hat{u}_B \):
\[ \hat{u}_B = \frac{\vec{B}}{|\vec{B}|} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} \]

### Step 4: Calculate the displacement vector for the second movement
Now, we calculate the displacement vector \( \vec{BC} \):
\[ \vec{BC} = 14 \cdot \hat{u}_B = 14 \cdot \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} = 2 \cdot (3\hat{i} - 2\hat{j} + 6\hat{k}) = 6\hat{i} - 4\hat{j} + 12\hat{k} \]

### Step 5: Calculate the total displacement vector
The total displacement vector \( \vec{AC} \) is the sum of \( \vec{AB} \) and \( \vec{BC} \):
\[ \vec{AC} = \vec{AB} + \vec{BC} = (18\hat{i} + 6\hat{j} + 9\hat{k}) + (6\hat{i} - 4\hat{j} + 12\hat{k}) \]
\[ \vec{AC} = (18 + 6)\hat{i} + (6 - 4)\hat{j} + (9 + 12)\hat{k} = 24\hat{i} + 2\hat{j} + 21\hat{k} \]

### Final Answer
The total displacement of the particle is:
\[ \vec{AC} = 24\hat{i} + 2\hat{j} + 21\hat{k} \text{ meters} \]