### Video Transcript

A particle moves in a plane in which π and π are perpendicular unit vectors. A force πΉ equals nine π plus π newtons acts on the particle. The particle moves from the origin to the point with position vector negative nine π plus six π metres. Find the work done by the force.

In this statement, weβre given a force πΉ and weβre told the components of that force. Weβre further told that the force helps to move a particle from the origin to the point negative nine π six π metres. Calling that position vector capital π, we want to solve for the work done by the force πΉ, which we can call π.

Letβs start out by drawing a diagram of the force and the position vector. If we draw in a set of coordinate axes, where π is the horizontal component and π is the vertical component, then we can graphically represent the force πΉ as well as the displacement π, which weβve drawn in moving from the origin to their respective component endpoints.

Since weβre solving for the work done π by the force πΉ, we can recall that that work is equal to the magnitude of the force πΉ times the magnitude of the displacement π multiplied by the cosine of the angle between those two vectors. In our case, we can write that π is equal to the magnitude of πΉ times the magnitude of π multiplied by the cosine of the angle between them, which we can call π.

We can start out by calculating what that angle π is. Looking at our diagram, if we call the angle between the positive π hat axis and πΉ πΌ and the angle between the negative π hat axis and π π½, then we can write that π is equal to 180 degrees minus πΌ minus π½ or converting from degrees to radians, π equals π minus πΌ minus π½.

Considering the angles πΌ and π½, we donβt know what they are, but we do know the endpoints of the vectors that start at the origin and help to define those two angles. The force vector πΉ ends at the point nine, one and the displacement vector π ends at the point negative nine, six. We can use these π, π coordinates of πΉ and π to help solve for πΌ and π½.

First, considering the angle πΌ, we can write that the tangent of πΌ is equal to one divided by nine. Thatβs the vertical component of that vector divided by its horizontal component. This indicates that πΌ is the inverse tangent of one-ninth.

Now, letβs consider the angle π½. We can write the tangent of π½ as positive six over nine β both positive because we can consider this as effectively being a first quadrant angle. So π½ equals the inverse tangent of six over nine. And we now have an expression for π that we can enter on our calculator and solve for. With that step done, letβs go back to our equation for work and consider the magnitude of the force and the magnitude of our displacement π.

Recalling that the magnitude of a vector is equal to the square root of the sum of the squares of its components, the components of πΉ are nine π and one π. So we take the square root of nine squared plus one squared. And that result will be equal to the magnitude of πΉ in units of newtons. And the magnitude of the displacement π will be equal to the square root of negative nine squared plus six squared, the squares of the components of π metres.

Bringing together our expressions for the magnitude of πΉ, the magnitude of π, and π, plugging these values in, and then entering them on our calculator, we find for a value of π negative 75 joules. Thatβs the work done on the particle by the force πΉ over its displacement π.