two force vectors f1=(3.0N)x- (3.0N)y and f2= (-6.0N)x + (4.5N)y are applied to a partical . what third force f3 would make make the net or resultant force on the partical zero?

add these two vectors f1 and f2. Then the negative of this resualant is the equilibrant.

To find the third force F3 that would make the net or resultant force on the particle zero, we need to add the vector components of F1, F2, and F3 in the x and y directions separately.

The net force in the x-direction (horizontal) will be zero when the sum of the x-components of the forces is zero. So, we can set up the equation:

F1x + F2x + F3x = 0

Substituting the given values, we have:

(3.0N) + (-6.0N) + F3x = 0

Simplifying the equation:

-3.0N + F3x = 0
F3x = 3.0N

Similarly, for the y-direction (vertical), the net force will be zero when the sum of the y-components of the forces is zero. So, we can set up the equation:

F1y + F2y + F3y = 0

Substituting the given values, we have:

(-3.0N) + (4.5N) + F3y = 0

Simplifying the equation:

1.5N + F3y = 0
F3y = -1.5N

Now, we have the x and y-components of the third force F3. The final answer would be the vector F3 = F3x * i + F3y * j, where i and j are the unit vectors in the x and y directions, respectively.

Therefore, F3 = (3.0N)i + (-1.5N)j or F3 = (3.0N, -1.5N) in component form.