Hey, folks. So in previous videos, we saw how to calculate the dot product between 2 vectors 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 going to have to calculate the dot product between 2 vectors using vector components instead. But what we're going to see in this video is that it actually works out to a pretty simple equation. So let's check it out. 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 2 magnitudes and then the cosine of the angle between them. So, basically, one way to think about this is that you're doing 3 and 4, the cosine of 60. We were calculating the component of \(b\) that is parallel to \(a\). So, basically, we just drew this little vector like this. This was my \(b \cdot \cos(\theta)\) or in other words, my \(4 \cdot \cos(60)\). This worked out to 2. So when we did this equation, we're actually multiplying 3 and 2 together, the parallel components, not 43, and then what we got was we just got the 6. So, sometimes, you're actually not going to be given magnitudes and angles and so you're going to have to use a different method instead using the vector components. A situation like \(2\mathbf{i} + 3\mathbf{j}\) multiplied by \(\mathbf{i} + 2\mathbf{j}\). So you have basically a bunch of vectors described by their unit vector components, which take the general form of \(a_x\) in the \(\mathbf{i}\) direction, \(a_y\) in the \(\mathbf{j}\), and \(a_z\) in the \(\mathbf{k}\) direction. Now the way that we're going to calculate the dot product in this way is using the same exact principle. So if you calculate the dot product between \(a\) and \(b\), you're just going to multiply the parallel components. And in these vectors here, they're organized by their \(\mathbf{i}\)'s \(\mathbf{j}\)'s and \(\mathbf{k}\)'s. The parallel components are just the \(x\)'s together, the \(y\)'s together and the \(z\)'s together. So that means that the formula just becomes \(a_x \cdot b_x\). You're multiplying the alike or parallel components. \(a_y\) and \(b_y\) and then you're doing \(a_z\) and \(b_z\). So you're 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 going to 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 going to use this equation here whenever you have diagrams and you can figure out the magnitudes and the angles between the vectors, and then you're going to use this equation here whenever you have the components of the vectors, usually \(\mathbf{i}\)'s, \(\mathbf{j}\)'s, and \(\mathbf{k}\)'s. Alright, guys. Let's get some practice. So we've got to calculate the dot product between these two vectors over here. So let's just get to it. If we want to calculate \(a \cdot b\), then all we have to do is just pair off the \(\mathbf{i}\)'s and then the \(\mathbf{j}\)'s together. So that means that \(a \cdot b\) is just going to be \(2 \cdot 1 + 3 \cdot 2\). So this just becomes 2 + 6 and that's 8. That's all there is to it. And notice how I just get a number out of this which is perfectly sensible because the scalar product should just get a number. So that's just 8. Alright. Let's do for the part \(b\). So now we're going to calculate the dot product between these. This is going to be \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\). This is going to be \(\mathbf{i}\) and \(\mathbf{j}\). So we're just going to pair off the components over here. So we've got my \(\mathbf{i}\)'s and then we've got the \(\mathbf{j}\)'s. Make sure to keep track of the signs over here. But now look at the \(\mathbf{k}\). The \(\mathbf{k}\) actually doesn't really have a pair. We're going to see how that works in just a second. So my \(a \cdot b\) is going to be Well, I've got \(-3 \cdot 1\). Right? So \(-3 \cdot 1 +\), then I've got \(\mathbf{i} \cdot (-2\mathbf{i})\). So don't forget there's a negative sign over there. So we've got \(1 \cdot (-2) +\), now you've got \(\mathbf{k}\) and then what's this pair over here? Well, this is a three-dimensional vector. This is a two-dimensional vector. So one way we can think about this is that the \(\mathbf{k}\) component is actually just 0. So when we do the dot products, we're just going to have \(4 \cdot 0\) and that term just goes away. That's really all there is to it. So that means my \(a \cdot b\) is just equal to \(-3 +\), and then this is going to be \(-2\), so I just get \(-5\). So that's the dot product. Alright, guys. That's all there is to it. Let's get some more practice.

# Calculating Dot Product Using Components - Online Tutor, Practice Problems & Exam Prep

### Calculating Dot Product Using Vector Components

#### Video transcript

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**).

85

15 i - 0.80 j - 12 k

11

57 i - 8.8 j + 37 k

### Calculating the Angle Between 2 Vectors Using the Dot Product

#### Video transcript

Hey, guys. Let's check out this example together. We've got these 2 vectors, a and b, and they're both written in terms of their unit vector components. So for the first part, we need to calculate a⋅b. So we've got our two forms of the dot product. We know the magnitude and direction, abcosθ. And we know the unit vector components where you just pair off all the parallel components and multiply and add them together. So which one are we gonna use? Well, we've got all these vector unit vector components. We got a bunch of i's and j'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⋅b. 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 I direction minus 3.9 in the j. And then for my b vector, I've got 2.1 in the I direction plus 4.8 in the j. And so if I want to calculate the dot product a⋅b, then I just pair off all the alike or the parallel components, the ones that point in the same direction and then you just multiply them. So, we're gonna pair off the i's, pair off the j's and then just multiply them and multiply them and add those pairs. So we've got 7.2, just the number, 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 or the x components together. And then these are the y components, so we multiply those together. And that is the dot products. That's just it. It's just -3.6. Remember, just to give you a number, and it could be positive or negative.

Okay. So let's move on to part b. Now in part b, we wanna calculate the angle that is between a and b. So for part b, now we're looking for an angle which you remember is just represented by the letter θ. So if you want θ between the two angles, which form of the equation are we gonna use? Well, remember how we said we use this for magnitude and direction and this for unit vector components? But remember that these two equations a⋅b are really just two ways to get to the same answer. You're still calculating ...