Following forces start acting on a particle at rest at the origin of the co-ordinate system simultaneously `vec(F)_(1)=-4hat(i)+5hat(j)+5hat(k), vec(F)_(2)=-5hat(i)+8hat(j)+6 hat(k), vec(F)_(3)=-3 hat(i)+4 hat(j)-7hat(k)` and `vec(F)_(4)=12hat(i)-3hat(j)-2hat(k)` then the particle will move-

Options

A. In x-y plane
B. In y-z plane
C. In x-z plane
D. Along x-axis

Explanation

To solve the problem, we need to find the net force acting on the particle by summing up all the forces given. The forces are:

1. \( \vec{F}_1 = -4\hat{i} + 5\hat{j} + 5\hat{k} \)
2. \( \vec{F}_2 = -5\hat{i} + 8\hat{j} + 6\hat{k} \)
3. \( \vec{F}_3 = -3\hat{i} + 4\hat{j} - 7\hat{k} \)
4. \( \vec{F}_4 = 12\hat{i} - 3\hat{j} - 2\hat{k} \)

### Step 1: Write down the forces
We have:
- \( \vec{F}_1 = -4\hat{i} + 5\hat{j} + 5\hat{k} \)
- \( \vec{F}_2 = -5\hat{i} + 8\hat{j} + 6\hat{k} \)
- \( \vec{F}_3 = -3\hat{i} + 4\hat{j} - 7\hat{k} \)
- \( \vec{F}_4 = 12\hat{i} - 3\hat{j} - 2\hat{k} \)

### Step 2: Sum the forces in the \( \hat{i} \) direction
Calculate the total force in the \( \hat{i} \) direction:
\[ F_{net, x} = -4 + (-5) + (-3) + 12 = 0 \]

### Step 3: Sum the forces in the \( \hat{j} \) direction
Calculate the total force in the \( \hat{j} \) direction:
\[ F_{net, y} = 5 + 8 + 4 + (-3) = 14 \]

### Step 4: Sum the forces in the \( \hat{k} \) direction
Calculate the total force in the \( \hat{k} \) direction:
\[ F_{net, z} = 5 + 6 + (-7) + (-2) = 2 \]

### Step 5: Write the net force vector
Now we can write the net force vector:
\[ \vec{F}_{net} = 0\hat{i} + 14\hat{j} + 2\hat{k} \]

### Step 6: Determine the direction of motion
Since the net force has components only in the \( \hat{j} \) and \( \hat{k} \) directions, the particle will move in the \( yz \) plane.

### Final Answer
The particle will move in the \( yz \) plane.

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