If vectors a ⃗ = 4 ı ^ a⃗=4î a ⃗ = 4 ı ^ and b ⃗ = 3 ı ^ − 2 ȷ ^ b⃗=3î-2ĵ b ⃗ = 3 ı ^ − 2 ȷ ^ , determine the angle between vectors a ⃗ a ⃗ a ⃗ and b ⃗ b ⃗ b ⃗ .

Let's try solving this problem. So here we're told if vector a is equal to 4 I, and vector b is 3 I minus 2 j, determine the angle between vectors a and b. So in essence, what we have going on are these 2 vectors. I'll go ahead and draw this graph over here. We'll give this an x axis and a y axis, and what we have is these two vectors. We have vector a, which is 4 I. So that means that we go 4 units to the right, and then this is gonna be vector a, and then vector b is 3 I minus 2 j. So this is gonna go 3 units to the right and 2 units down. And, by the way, I'm just kinda drawing these vectors that look like about what we should have. It's not exactly because we don't have any numbers on here, but this is what the 2 vectors are. And And what we're trying to do is figure out what this angle is between the 2 vectors. So that's the angle that we're looking for, and we can actually calculate this using an equation that we've learned about. The equation is that the dot product of a and b is going to equal the magnitude of a times the magnitude of b times the cosine of the angle between those two vectors. Now there's a little bit of a problem with this equation that we have right now though. We don't know what the dot product of a and b is, nor do we know what these magnitudes are. So these are all things that we're going to have to find. Now to start things off, I'm going to find the dot product of a and b, and we should be familiar with the dot product operation. What we can do is take our first vector, which is 4 I, and we can find its dot product with the second vector, which is 3 I minus 2 j. And recall that when doing the dot product, what you want to do is multiply like components and then add them together. So So we're going to multiply this 4 and this 3, which will give us 12. And then we're going to add that to the dot product with the j components, but notice that we don't have another j component over here. So basically, we have 0 times 2, which comes out to just 0, so our dot product is 12. So this is the dot product for the 2 vectors a and b. Now we also need to figure out what the magnitudes of a and b are. Well, to find the magnitude of a, what I can do is use the equation where we take the square root of the x component squared, plus the y component squared. Now the x component is 4 because we have 4 I, and then the y component would be the the j, but we don't have any j's up here. So all we're going to have is the square root of the x component, which is 4 squared. And then since we don't have any y components, we don't have any j's, we're just gonna say that that's 0 squared. So the square root of 4 squared is just 4, so that's the magnitude of a. Now we also need to find the magnitude of b. To find the magnitude of b, which I'll do over here, well, that's going to be the square root of 3 squared minus or a plus, I should say, negative 2 squared, because we see that the x component is 3 and the y component is negative 2. Now 3 squared is 9, and negative 2 squared is positive 4. And 9 +4 is 13, so we end up with the square root of 13. So this right here is the magnitude of b. This right here is the magnitude of a, and now we have the dot product, so we can now use this equation. So we're going to have that this whole thing is the dot product of a and b, which we can see here is 12, and that's going to be equal to the magnitude of a, which is 4, times the magnitude of b, which is the square root of 13 times the cosine of the angle between these vectors. Now from here, all I need to do is solve for the cosine of theta, which I'll do down there. So I'm first going to divide 4 times the square root of 13 on both sides of this equation. This is going to get 4 times the square root of 13 to cancel on the right side, leaving us with just the cosine of theta. And the cosine of theta is going to be equal to this fraction, which is 12 divided by 4 times the square root of 13. Now you can actually reduce this fraction because 12 divided by 4 will give you 3. So we can actually write this as 3 divided by the square root of 13. Now you could rationalize the denominator if you want to, but since we're gonna just plug this number into a calculator, I'm actually just gonna leave it like for this video. So what I'm gonna do is take the inverse cosine on both sides of this equation, that's going to get the cosine to cancel on the left side, leaving us with just theta, and we'll have that theta is equal to the inverse cosine of this fraction right here. So it's gonna be the inverse cosine of 3 divided by the square root of 13. Now this is something that you can plug into a calculator, and if you do this, you should get assuming you're in degree mode, that your inverse cosine of 3 divided by the square root of 13 is approximately equal to 33.69 degrees. So this right here would be the vector or, excuse me, the angle between vectors a and b. So this angle right here that we were talking about, that's 33.69 degrees approximately between the two vectors. So hope you found this video helpful. Thanks for watching.