# How to Do Vector Operations Using Components Video

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• 0:11 Vector Components
• 1:24 Breaking a Vector into…
• 3:44 Example Problem
• 5:27 Lesson Summary
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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this lesson, you will understand how we analyze physics involving two-dimensional vectors and be able to split a vector at an angle into a vertical and horizontal component. A short quiz will follow.

## Vector Components

In physics, we love to keep things simple. But not everything goes in nice, neat straight lines. When you shoot a cannonball at a pirate ship, that pirate ship moves in at least two dimensions -- the x-direction and the y-direction. But so many of our physics equations only work in one dimension. So what do we do?

We break vectors into x and y components. We look at the motion in the x-direction and the motion in the y-direction separately.

A vector is a quantity that has both magnitude (numerical size) and direction. So for example, 20 meters per second is not a vector and is called speed, but 20 meters per second south is a vector and is called velocity. If that vector is pointed diagonally, to analyze the motion, we need to break it into vertical (y) and horizontal (x) components.

If your cannonball is shot at a shallow angle, close to the ground, you're going to have a bigger x-component and a smaller y-component. If your cannonball is shot almost upwards, you're going to have a bigger y-component and a smaller x-component. And we can put numbers to these things and treat the motion in the two dimensions (or three dimensions for that matter) separately. This is incredibly helpful for solving problems.

## Breaking a Vector into Components

Okay, so how do we do it? We use a bit of geometry. The word that you may have heard before is: SOHCAHTOA. This helps us find the sides of a right-angled triangle when we know one of the sides and we know an angle. The SOH part reminds us that sine of the angle is equal to the opposite side of the triangle divided by the hypotenuse side of the triangle. The CAH part tells us that the cosine of the angle is equal to the adjacent side of the triangle divided by the hypotenuse. And the TOA part says that the tangent of the angle is equal to the opposite side of the triangle divided by the adjacent.

As you can see from this graphic, the opposite side is the side directly opposite from the angle that you're given. The hypotenuse is the longest side, the diagonal, which is always directly across from the right angle. And the adjacent side is the side that is adjacent to, or next to, both the right angle and the angle you're given.

Okay, but this is physics, not math. So why do we need to know that?

Well, if you're shooting something up at an angle, or have any kind of physics vector, like a force vector, electric field vector, or whatever it is. If you draw it to scale, you can create a right-angled triangle from that vector. One side of the triangle should point directly in the x-direction -- perfectly horizontal. The other side should point directly in the y-direction -- vertical. And if you know the angle it's being shot at, you can use this SOHCAHTOA geometry to find the x and y components of the velocity, or force, or electric field, or whatever it is.

So for example, with this cannonball being fired at an angle, we're firing it at a velocity v, that's the hypotenuse, and that has an x-component, vx, and a y-component, vy. We can write a couple of equations using SOHCAHTOA. Or really just SOHCAH, because we only use sine and cosine for breaking into components.

We can write an equation using sine. Sine of the angle is equal to the opposite side, which is vy, divided by the hypotenuse, which is v. And we can also write an equation using cosine. Cosine of the angle is equal to the adjacent side, which is vx, divided by the hypotenuse, which is v. Rearrange these equations, and we get expressions for the x and y components of the velocity, v.

So that's how we find our two components. But maybe this would be easier if we looked at an example problem.

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