An insect starts from `(2,3,4)` and travels along `(hat(i)+2hat(j)+2hat(k))` with `15 m//s` for `2sec`. Then it reverse it direction and travels with `3m//s` for `5` sec. If final position of the insert is `(x,y,z)`. Then find the value of `(x+y+z)` ?

Options

A. `34 m`
B. `35 m`
C. `36 m`
D. `63 m`

Explanation

To solve the problem step by step, we will follow the journey of the insect from its starting position to its final position.

### Step 1: Determine the initial position of the insect
The insect starts from the position given by the coordinates:
\[ (2, 3, 4) \]

### Step 2: Calculate the distance traveled in the first leg of the journey
The insect travels in the direction of the vector \((\hat{i} + 2\hat{j} + 2\hat{k})\) at a speed of \(15 \, \text{m/s}\) for \(2 \, \text{s}\).

First, we calculate the distance traveled:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 15 \, \text{m/s} \times 2 \, \text{s} = 30 \, \text{m} \]

### Step 3: Find the unit vector of the direction
Next, we need to find the unit vector in the direction of \((\hat{i} + 2\hat{j} + 2\hat{k})\).

First, calculate the magnitude of the vector:
\[ \text{Magnitude} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Now, the unit vector \(\hat{u}\) in the direction of \((\hat{i} + 2\hat{j} + 2\hat{k})\) is:
\[ \hat{u} = \frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \]

### Step 4: Calculate the position after the first leg of the journey
Now we can find the new position after traveling \(30 \, \text{m}\) in the direction of the unit vector:
\[ \text{New Position} = (2, 3, 4) + 30 \left(\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right) \]
Calculating this gives:
\[ = (2, 3, 4) + (10, 20, 20) = (12, 23, 24) \]

### Step 5: Calculate the distance traveled in the second leg of the journey
The insect then reverses its direction and travels at \(3 \, \text{m/s}\) for \(5 \, \text{s}\):
\[ \text{Distance} = 3 \, \text{m/s} \times 5 \, \text{s} = 15 \, \text{m} \]

### Step 6: Calculate the new position after reversing direction
Since the insect is reversing its direction, we need to subtract the distance traveled in the direction of the unit vector:
\[ \text{New Position} = (12, 23, 24) - 15 \left(\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right) \]
Calculating this gives:
\[ = (12, 23, 24) - (5, 10, 10) = (7, 13, 14) \]

### Step 7: Find the final coordinates
The final coordinates of the insect are:
\[ (x, y, z) = (7, 13, 14) \]

### Step 8: Calculate \(x + y + z\)
Now, we can find the value of \(x + y + z\):
\[ x + y + z = 7 + 13 + 14 = 34 \]

Thus, the final answer is:
\[ \boxed{34} \]