A particle starts moving from the point (1, 2, 3) under velocity ` 1 hat I +2hat j + 3hatk ` for two seconds. After that it moves under velocity ` 1 hat I -2hat j + 3hatk ` for the next two seconds. What is the coordinate of the final points?

Options

A. ` (5,2,5) `
B. ` ( 5,2,15) `
C. `( 1,2,15)`
D. none of these

Explanation

To solve the problem step by step, we need to calculate the position of the particle at different stages of its motion.

### Step 1: Initial Position
The particle starts from the point \( (1, 2, 3) \). We can represent this in vector form as:
\[ \mathbf{S_0} = 1 \hat{i} + 2 \hat{j} + 3 \hat{k} \]

### Step 2: First Velocity and Time
The particle moves with a velocity of \( \mathbf{V_1} = 1 \hat{i} + 2 \hat{j} + 3 \hat{k} \) for \( t_1 = 2 \) seconds. The displacement during this time can be calculated using:
\[ \mathbf{S_1} = \mathbf{V_1} \cdot t_1 = (1 \hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot 2 = 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \]

### Step 3: Position After First Interval
Now, we find the new position after the first interval:
\[ \mathbf{S_{total1}} = \mathbf{S_0} + \mathbf{S_1} = (1 \hat{i} + 2 \hat{j} + 3 \hat{k}) + (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) = (1 + 2) \hat{i} + (2 + 4) \hat{j} + (3 + 6) \hat{k} = 3 \hat{i} + 6 \hat{j} + 9 \hat{k} \]

### Step 4: Second Velocity and Time
Next, the particle moves with a new velocity \( \mathbf{V_2} = 1 \hat{i} - 2 \hat{j} + 3 \hat{k} \) for \( t_2 = 2 \) seconds. The displacement during this time is:
\[ \mathbf{S_2} = \mathbf{V_2} \cdot t_2 = (1 \hat{i} - 2 \hat{j} + 3 \hat{k}) \cdot 2 = 2 \hat{i} - 4 \hat{j} + 6 \hat{k} \]

### Step 5: Final Position After Second Interval
Now, we find the final position after the second interval:
\[ \mathbf{S_{final}} = \mathbf{S_{total1}} + \mathbf{S_2} = (3 \hat{i} + 6 \hat{j} + 9 \hat{k}) + (2 \hat{i} - 4 \hat{j} + 6 \hat{k}) = (3 + 2) \hat{i} + (6 - 4) \hat{j} + (9 + 6) \hat{k} = 5 \hat{i} + 2 \hat{j} + 15 \hat{k} \]

### Final Answer
Thus, the coordinates of the final point are:
\[ (5, 2, 15) \]

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