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Calculating Dot Product Using Vector Components

Patrick Ford
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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.