3. Vectors

Calculating Dot Product Using Components

# Calculating the Angle Between 2 Vectors Using the Dot Product

Patrick Ford

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Hey, guys, let's check out this example. Together, we've got these two vectors A and B, and they're both written in terms of their unit vector components. So for the first part, we calculate a dot Be so we've got our two forms of the dot product. We know the magnitude and direction A b cosine, theta. We know the unit vector components where you just pair off all the parallel components and multiplying album together. So which one we gonna use? What? We've got all these vectors, Unit vector components. We got a bunch of eyes and Jay's so we're gonna use this form here to calculate the dot product. Let's get to it. So this first part, part A we're gonna calculate a dot Be So first we have to just write out each equation or each of the vectors. We're gonna stack them on top of each other. So we've got 7.2 in the eye direction minus 3.9 and the J and then for my B vector, I've got to one in the eye direction plus four a in the J. And so if I want to calculate the dot product a dot be. Then I just pair off all the alike or the parallel components, the ones at the point in the same direction. And then you just multiply them. So we're gonna pair off the eyes, pair off the jays, and then just multiply them and add those pairs. So we've got 7.2 just the number of times 2.1, and then we've got plus negative, 3.9 times 4.8. So notice how we've just multiplied all the I components together, the x components together. And then these are the white components. We multiply those together like this. Okay, so now all we do is just plug this into our calculators and you should just get negative 36 And that is the dot products. That's just it. It's just negative. 3.6, remember? Just give you a number on it. Could be positive or negative. Okay, so let's move on to part B now, in partly what To calculate the angle that is between A and B. So for part B. Now we're looking for an angle which remember is just represented by the letter Thatta. So if you want data between the two angles. Which form of the equation we're gonna use won't remember how we said we use this for magnitude and direction and this for unit vector components. But remember that these two equations a dot b are really just are these two equations with two forms are really just two ways to get to the same answer. You're still calculating a dot be you're just doing it using the unit vector components. But sometimes you might actually have to go between the equations and get to the other one. For example, we want the we already calculated with the DOT product is by using this equation over here, which has got negative 3.6. But now we actually want to go get the angle, which is the cosine theta term that is, between these two vectors over here A and B. So we're gonna have to use this form of the equation. So we want to use that a dot B which, by the way, we already know what this is. Remember, this is just 3.6 is equal to the magnitude of a times the magnitude of B times, the co sign of the fate of term. So this is really what we're trying to do is really, really what we're trying to look for. So if we already have what the dot product is, the only thing that's left to figure out is what the magnitudes of the vectors are, right? We just have to figure out what the magnitude of A and B are. So sometimes we're just giving in the diagram here. We actually just have to calculate it on the way we can calculate it just by using the unit vector components. So if I want to figure out the magnitude of a remember, I already have the unit vector components over here, so I'm just gonna use my Pythagorean theorem member. This is just the legs of the triangle. So I've got 7.2 squared, plus negative +39 squared and you just get eight to we do the same thing for B Magnitude is just gonna be the Pythagorean theorem. Um, so we're gonna use these two things over here, So let's just say all that stuff goes here. So we've got 2.1 squared plus 4.8 square, and we're gonna get 5.2, so These are the two magnitudes. So now we could just plug these numbers back into our equation here and then figure out what that CoSine theta term is. So I've got negative. 3.6 equals five hoops equals 8.2 times 5.2 times the coastline of the state of term. So I could just move this stuff over. And when you divide that stuff well, you got negative. 3.6 divided by 8.2 times 5.2. And if you plug this into your calculator, what you should get is you should get the coastline. If they did, term is equal to negative zero point 08 Whoops. There we go. Negative 0.8 So now this is my data term. Remember that this data still kind of locked up inside of this cosine term. But I know that the coastline of data is equal to this number over here. So if I wanted to sort of extract or get that coast, get that data term outside of this, uh, this co sign, What I do is I take the inverse cosine. So if cosine theta is negative 0.8 then Data is just the arc co signer, the inverse cosine off negative 0.8 So if you do that, you're actually gonna get 94.6 degrees. And that is the answer for the angle between these two vectors. One way you can see this really, really quickly is if we kind of just sketch out what each of the vectors look like. So we know the A vector is gonna be 7.2 minus 3.9. So it's gonna look something like this in this direction. So it's gonna have a positive X in a negative light component, and then my be vector is gonna be 2.1 plus 4.8. So it's actually gonna be it's gonna look something like this. It's gonna look really close to a to a right angle. And in this sketch, it's kind of hard to see. So that's actually why we got a number that's really, really close to 90 degrees. Um, it just happens to be a little bit past 90 degrees. So that's why we got a slightly negative components. So anyway, so that's the sketch. Let me know if you guys have any questions. That's it for this one

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