For the two vectors A and B in Fig. E1.39,

Question

find (b) the magnitude and direction of the vector product A x B

Accepted Answer

everyone in this video. We're asked to use the information in the image to determine the magnitude and direction. For the vector product of M and n. Okay. We hear the term vector product. We can also think of that as the cross product. Okay. Were given the vectors M. M is 4.2 m And it's 52° above the x axis. Then we have N. And is in the third quadrant and it is 28 degrees below the Y axis and its length is 3.2 m. Alright, So when we're asked to find direction and magnitude, let's go ahead and start with the direction. Okay, so for the direction we want to use our right hand rule. So we take our hand. We pointed in the direction of em with our palm facing the direction of N. Ok. We're gonna curl our fingers towards the vector N. Ok. And we're doing it in the shortest possible distance. So our fingers are pointing towards em. We're curling towards end and you'll see that your thumb is pointing towards you out of the page. Okay. And what that means. Okay, so thumb points out what that means is that we're in the positive Z direction. I'm pointing out. Okay, The positive Z direction. Okay, so that's what we need to do to find the direction. And now for the magnitude while we know we can write the magnitude of C. We'll call our vector product C. Okay, so that's gonna be victor. M crossed with vector. And we can recall that we can write that as the magnitude of the vector M times the magnitude of the vector. N. All times sine data. Okay. And data is going to be the angle between M and N. And it's going to be the smallest possible angle you can find. Okay, that goes between them. So we aren't going to go to this angle down below because that would be larger. We're gonna go to this angle Theta. Hear from em all the way over here to Okay, that's going to be our theta. And what we're going to see is that data is made up of three things. Okay, we have theta one here that goes from em to the y axis. We have theta two here that goes from the positive y axis to the negative X axis. Data square data to And we have finally data three. That goes from the negative X axis down to vector N. This is data three. Okay. So we have three different vector components that we need to determine. Okay. And this one will just label with blue. Alright, so our theta is going to be equal to theta one. Plus data to plus data three. Okay, so that big angle theta between vectors M and N is gonna be made up of those three components. Let's write that out. Well, data one. We know that this angle below is 52°. Okay. And we're in a 90° angle. So we know that data one is going to be 90 degrees minus the angle 52. Okay. Data to well, it's the entire quadrant. So that's gonna be 90°. And data three is given. That's the angle between the X axis, the negative X axis and our vector. And that's 28 degrees. Okay. So adding those up, we are going to have 156°. All right. So then if we go back to our cross products, see the magnitude of that vector product. We have The magnitude of vector m well, that's 4.2 meters With the magnitude of Vector N. That's 3.2 meters. And then we have sine of our angle theta which we just found to be 156°. Alright, so working that out, we are going to get the magnitude of C. Is 5.47. We have meters times meters. So we have meters squared. So there we have it. We have our magnitude see 5.47 m squared. We found our direction using the right hand rule positive Z direction. Okay. And this is going to correspond with answer D. I hope that video helped you. Thanks for watching

For the two vectors A and B in Fig. E1.39, find (b) the magnitude and direction of the vector product A x B

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For the two vectors A and B in Fig. E1.39,
find (b) the magnitude and direction of the vector product A x B