Find the vector product A x B (expressed in unit vectors) of the two vectors given in Exercise 1.38. What is the magnitude of the vector product? Given two vectors A = 4.00 i + 7.00j and B = 5.00 i − 2.00

Video transcript

Hey everyone in this problem, we are given two vectors M equals five, I plus four J. And n is equal to three I minus seven J. And we're asked to find the vector product M cross N. In unit vectors ask what the magnitude of the vector product is as well. So when we're talking about vector product M Cross N. Okay, you can call this the vector product or the cross product. If you're more familiar with that term now let's start by getting an idea of what direction this is going to be going. Okay, we need the magnitude and the direction. So if we use our right hand rule, okay, we point our fingers in the direction of em we curl them in the direction of N. And our thumb is going to point in the direction of M cross N. And our thumb is going to be pointing into the page. Okay, or in the negative z direction. Alright. So automatically, if you look at the answer choices, there's only one that has a direction in the negative Z direction. So we would expect that we're going to get answer B But let's go ahead and find the magnitude of this vector product and just be sure. Okay, so what we can do is we can go ahead and use the determinant of a three by three matrix in order to calculate the vector product. And what does that matrix looks like? Well, we want to find the determinant of the matrix whose first row is I hat J hat and ke ha, okay, representing those three directions. The second row is going to be the corresponding values for our vector. M. So in the I hat direction we have five units in the j hat direction we have four units. And in the chaos direction we have zero units. Similarly for n that's going to be the final row. We have three in the direction, -7 in the Jihad direction and zero in the Chaos direction. Okay, so we have this three by three matrix. We want to take the determinant of it and let's go ahead and use Cramer's rule. If we use Cramer's rule and we take this last column as the column to expand, then we're going to end up with zeros. Okay, so let's do that. So we're going to take the first element in the last row, K hat and we're going to multiply it by the determinant of the two by two matrix made when we eliminate the row and column that catering. So if we eliminate row one, column three, we're left with 543 negative seven as our two by two matrix. And then we're gonna subtract The next element in that row or in that column which is zero times the determinant of the matrix made when you eliminate the row and column that that element is in. So we eliminate row two column three. We're left with I hat J hat in the first row, 3 -7 in the 2nd row. And finally we add The final element zero times the determinant of the two by two matrix made by eliminating that row and column. So the third row, third column. And we're left with I hat J hat in the first row and 54 in the second row. Okay. And you'll notice that these second these last two terms both have zero times some determinant. So those are just going to be zero. So we don't even need to worry about those determinants and we're gonna end up with a hot times. Now the determinant of a two by two matrix. We start with the first element, first row, first column. We multiply the diagonal. So we have five times negative seven. And then we're gonna subtract the product of the opposite diagonal three times 4. So this is gonna be k hat times negative 35 minus 12, which is equal to negative 47 K hat. And this is going to be M cross N. Okay. M cross N is equal to the determinant of this matrix. And so we have the cross product or the vector product, M. Cross N is negative 47 K hat. Okay. And so we have no I or J component. We only have a K component. So we know we're in the Z direction and this is negative. So we are indeed in the negative Z direction. Just like we expected using our right hand rule. Okay. And if we take the magnitude of this cross product, M cross N. This is gonna be the square root of the square of these components. Well, the I had component is zero, The J. Hat component is zero And the K Hot component is -47. And when we do this we get a magnitude of 47. And so the magnitude of our cross product of vector product is 47. It's in the negative Z direction, which we found both by calculating the vector product and by using our right hand rule. And so we have answer choice B. That's it for this one. Thanks everyone for watching. See you in the next video.