# Dot Product stuff ;-(

• PsychonautQQ
In summary, the work done on the particle can be found by taking the dot product of the force and displacement vectors. The angle between the vectors can be found using the formula above.

## Homework Statement

A constant force of 1i - 5j -8k moves (1,-4,2) (-3,2,-1), what is the work done on the particle?

## Homework Equations

Avector*Bvector=ABsinθ
?? I think?

## The Attempt at a Solution

I really am quite lost... but I found the coordinates for the position vector...

(-3-1)i + (2--4j) + (-1-2)k and the magnitude of the position vector is 7.81
so the unit vector is -.512i + .768j -.384k... am I on the right track here?

PsychonautQQ said:

## Homework Statement

A constant force of 1i - 5j -8k moves (1,-4,2) (-3,2,-1), what is the work done on the particle?

## Homework Equations

Avector*Bvector=ABsinθ
?? I think?

## The Attempt at a Solution

I really am quite lost... but I found the coordinates for the position vector...

(-3-1)i + (2--4j) + (-1-2)k and the magnitude of the position vector is 7.81
so the unit vector is -.512i + .768j -.384k... am I on the right track here?

For a constant force F, the work done is just the dot product of the force with the displacement. Use that.

so the force is 1i - 5j -8k and the displacement is -4i +6j -3k? so dot product would be -4 -30 - 24 = -58?

How do I find the angel between these vectors?

PsychonautQQ said:
so the force is 1i - 5j -8k and the displacement is -4i +6j -3k? so dot product would be -4 -30 - 24 = -58?

Yes, if the force isn't constant you need to work harder and integrate, but if it's constant, it's that easy.

PsychonautQQ said:
How do I find the angel between these vectors?

Why do you think you need the angle?

If you are required, separately, to find the angle between vectors, the dot product can be defined as
$$\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$$

where $\theta$ is the angle between the vectors.