Following forces start acting on a particle at rest at the origin of the co-ordinate system
`overset(rarr)F_(1)=-4 hat I -5 hat j +5hat k , overset(rarr)F_(2)=5hat I +8 hat j +6hat k , overset(rarr)F_(3)=-3hat I + 4 hat j - 7 hat k` and `overset(rarr)F_(4)=2hat i - 3 hat j - 2 hat 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 resultant force acting on the particle by summing the individual forces. The forces given are:

1. \( \overset{\rarr}{F_1} = -4 \hat{i} - 5 \hat{j} + 5 \hat{k} \)
2. \( \overset{\rarr}{F_2} = 5 \hat{i} + 8 \hat{j} + 6 \hat{k} \)
3. \( \overset{\rarr}{F_3} = -3 \hat{i} + 4 \hat{j} - 7 \hat{k} \)
4. \( \overset{\rarr}{F_4} = 2 \hat{i} - 3 \hat{j} - 2 \hat{k} \)

### Step 1: Sum the Forces

We will sum the forces component-wise.

**I-component:**
\[ F_{net_x} = -4 + 5 - 3 + 2 \]
Calculating this gives:
\[ F_{net_x} = 0 \]

**J-component:**
\[ F_{net_y} = -5 + 8 + 4 - 3 \]
Calculating this gives:
\[ F_{net_y} = 4 \]

**K-component:**
\[ F_{net_z} = 5 + 6 - 7 - 2 \]
Calculating this gives:
\[ F_{net_z} = 2 \]

### Step 2: Write the Resultant Force

Now we can write the resultant force vector:
\[ \overset{\rarr}{F_{net}} = 0 \hat{i} + 4 \hat{j} + 2 \hat{k} \]

### Step 3: Analyze the Resultant Force

The resultant force has:
- No component in the x-direction (\(0 \hat{i}\))
- A positive component in the y-direction (\(4 \hat{j}\))
- A positive component in the z-direction (\(2 \hat{k}\))

### Step 4: Determine the Motion of the Particle

Since the resultant force has components in the y and z directions but no component in the x direction, the particle will move in the yz plane.

### Conclusion

The particle will move in the yz plane.