Calculating Dot Product Using Components - Video Tutorials & Practice Problems

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concept

Calculating Dot Product Using Vector Components

Video duration:

4m

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Hey, folks. So in previous videos, we saw how to calculate the dot product between two vectors by using their magnitudes and angles a situation like this where we had A and B and the cosine of the angle between them. And so in some problems, you're gonna have to calculate the dot product between two vectors by using vector components instead. But what we're gonna see in this video is it actually works out to a pretty simple equation. So let's check it out. So, guys, remember that the dot product is the multiplication of parallel components. For example, when we did this with magnitudes and angles, we used a simple formula, which is the multiplication of the two magnitudes and then the coastline of the angle between them. So basically, one way to think about this is that you're doing three and four. The coastline of 60. We were calculating the components of B that is parallel to a so basically just drew this little vector like this. This was my be co sign of data or, in other words, my four times the cosine of 60. This worked out to two. So when we did this equation we're actually multiplying three and two together. The parallel components, not four and three. And then what we got was we just got six. So sometimes you're actually not gonna be given magnitudes and angles, and so you're gonna have to use a different method instead using the vector components a situation like to I and three j multiplied by I and two J. So you have basically a bunch of vectors described by their unit vector components, which take the general form of a X in the eye direction a Y in the J and a Z in the K direction. Now the way they were going to calculate the DOT product in this way is using the same exact principle. So when if you calculate the dot product between A and B, you're just gonna multiply the parallel components and in these vectors here that are organized by their eyes Jason case, The parallel components are just the exes together with wise together and the Z's together. So that means that the formula just becomes a X Times bx. You're multiplying the alike or parallel components a y and B y, and then you're doing ese and busy So you just pairing off each one of these little parallel components and then you're just adding them all together. So you just do this. These are all just gonna be numbers like this. And then that's really all there is to it. That's your dot product. Alright, guys, that's really all there is to it. So you're gonna use this equation here whenever you have diagrams and you can figure out the magnitudes and the angles between the vectors and you're gonna use that This equation here, whenever you have the components of the vectors, usually eyes Jason case. Alright, guys, let's get some practice. So we've got to calculate the, uh, the dot product between these two vectors over here. So let's just get to it if we want to calculate a dot B and all we have to do is just pair off the eyes and then the jays together. So that means that a dot be is just gonna be too times one plus three times to. So this just becomes two plus six and that's eight. That older is to it. And so notice how I just get a number out of this, which is perfectly, which which makes sense, because the Scaler products should just get a number. So that's just eight. Alright, let's do for the party. So now we're gonna calculate the dot product between these. This is gonna be I, j and K. This is gonna be I and J. So we're just gonna pair off the components over here. So we've got my eyes and then we've got the jays. Make sure to keep track of the signs over here. But now look at the K. The K actually doesn't really have a pair. We're gonna see how that works in just a second. So my a dot B is gonna be Well, I've got negative three times. One rights of negative three times one plus then I've got I times negative too. I so don't forget there's a negative sign over there. So you've got one times negative. Two plus, Now we've got K. And then what's this? What's this pair over here? Well, this is a three dimensional vector. This is a two dimensional vector. So one we can think about this is that the K component is actually just zero. So when we do the dot product we're just gonna have four times zero. And that term just goes away. That's really all there is to it. So that means my a dot B is just equal to negative three plus. And then this is gonna be negative too. So I just get negative five. So that's the dot product. Alright, guys, that's all there is to it. Let's get some more practice.

2

Problem

Problem

Calculate the dot product between A = (6.6 i - 3.4 j - 6.4 k ) and B = (8.6 i + 2.6 j - 5.8 k).

A

85

B

15 i - 0.80 j - 12 k

C

11

D

57 i - 8.8 j + 37 k

3

example

Calculating the Angle Between 2 Vectors Using the Dot Product

Video duration:

5m

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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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