Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Finding Angle Using Scalar & Vector Products

Patrick Ford
Was this helpful?
Hey guys, So let's get started with our problem here. So we've got these two vectors A. And B. And we're told some information about their scalar and vector products there dot and cross products and we want to calculate what their angle between them is. So in other words, what we're asked to find here is theta. But if you notice, I actually know nothing about these vectors, I don't not given a graph, I don't know what their magnitudes and directions or anything like that. So how do we figure this out? Well, if you're looking for theta, the one equation that pops up here with the theta term is going to be that the magnitude of the vector products, remember this really just produces another vector. Just going to call it C. Is equal to a B. Sine theta, the two magnitudes and the angle between them. So really this is our target variable and we actually know what this magnitude is, right? Usually we're asked to find it, but we actually know it's already 12. So if I try to rearrange this equation, what I'm gonna end up with is that the sine of theta is equal to 12 over a B. And I can't really do anything else with this, right? So I can't even if I were to try to do this inverse sine, I still don't know what this A and B are. So, because I don't know what the magnets of those vectors, I can't really finish off this equation, I'm going to need something else. I'm gonna need another equation. Well, the one thing that we haven't seen that we haven't used yet, is that the scalar product is negative eight. Remember the scalar product is A dot B. So what I'm gonna do here is I'm gonna start out another equation and I'm gonna say that a dot B is equal to remember that the definition of a dot B. It doesn't have a magnitude, it's just a number and it was just a B. Cosine theta. Except we already know what the scalar product is. We know that A dot B. Whatever. Whenever you do A B. Cosine, theta, you're gonna get negative eight. Alright, so here's what I'm gonna do, I'm gonna rewrite another equation. I'm gonna get cosine of theta, right? Because this is my other, this is my target variable, just in another equation. And this is going to be -8 divided by a B. So here's what I'm gonna do, right? I've got these two equations and they actually both have three unknowns. I've got the theta that's unknown and I've got my A and B. That's unknown. So what I can do to get rid of them is I can actually divide the two equations. So what I'm gonna do here is I'm gonna do sine of theta divided by cosine of theta. And what you're going to get here is that 12 divided by A B divided by negative eight Over A B. And what happens here is that the A B. S are going to cancel when you divide these two equations, the abs cancel out. And then basically what you end up with here is just a tangent. Theta. Right? Sine over cosine equals tangent. So you've got the tangent, theta is equal to and this is just 12 over negative eight. So this is just gonna work out to negative 1.5. So we've got these two equations that had three unknowns and by dividing them we were actually able to get down to only one equation with one unknown. So now what I have to do is just I just take the inverse tangent of this and just make sure that your calculator is in degrees. You're gonna take the inverse tangent of negative 1.5. What you're gonna get here is you're gonna get negative 56 points. So you're gonna get like negative 60 56.3 degrees. Alright. So is that our final answer? Well remember that this angle here, this theta is the angle between these two vectors A and B. But it has to be the smallest positive value. So what we can do here is if this angle here is negative 56 then we're gonna have to add 100 and 80 degrees to it. one way you can kind of think about this Um is that these two vectors have a scalar product of -8. So what happens here is I'm just gonna draw like a like a sketch real quickly, what these vectors might look like. So this is a sketch. So this is my X and Y axis. So, if this vector is like, let's say this is a then in order for them to have a scalar product that's negative, It means that they have to be pointing in opposite directions. You only get scalar products that are negative whenever you have some components that point anti parallel. So basically what happens is we know that the angle between A and B has to be greater than 90 degrees. So this data here must be Greater than 90° because a dot B is negative is less than zero. Alright, So now that we've added 182 this what you're gonna get is 100 and 23.7 degrees and that is the right answer. All right. So now we know that these angle this angle here has to be 123.7. That's the only way you can get A vector product that has a magnitude of 12. But a scalar product that has a magnitude Or that's a scalar product, that's negative eight. Alright, so hopefully that makes sense. Let me know, guys, if you guys have any questions and I'll see in the next one