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 net force acting on the particle by summing up all the forces given. The forces are represented as vectors in a three-dimensional coordinate system. Let's break down the solution step-by-step.

### Step 1: Identify the Forces
We are given four forces acting on the particle:
- \( \vec{F_1} = -4 \hat{i} - 5 \hat{j} + 5 \hat{k} \)
- \( \vec{F_2} = 5 \hat{i} + 8 \hat{j} + 6 \hat{k} \)
- \( \vec{F_3} = -3 \hat{i} + 4 \hat{j} - 7 \hat{k} \)
- \( \vec{F_4} = 2 \hat{i} - 3 \hat{j} - 2 \hat{k} \)

### Step 2: Calculate the Net Force
The net force \( \vec{F_{net}} \) can be calculated by adding all the forces together:

\[ \vec{F_{net}} = \vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} \]

### Step 3: Add the Components
Now, we will add the components of each vector separately (i.e., the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) components).

#### i-component:
\[ F_{net, i} = -4 + 5 - 3 + 2 = 0 \]

#### j-component:
\[ F_{net, j} = -5 + 8 + 4 - 3 = 4 \]

#### k-component:
\[ F_{net, k} = 5 + 6 - 7 - 2 = 2 \]

### Step 4: Combine the Results
Now we can write the net force vector:

\[ \vec{F_{net}} = 0 \hat{i} + 4 \hat{j} + 2 \hat{k} \]

### Step 5: Determine the Direction of Motion
Since the net force has a component in the \( \hat{j} \) and \( \hat{k} \) directions but no component in the \( \hat{i} \) direction, the particle will move in the direction of the net force, which is primarily in the \( \hat{j} \) and \( \hat{k} \) directions.

### Conclusion
The particle will move in the direction of the net force, which lies in the \( y-z \) plane (since the \( x \)-component is zero).