If vectors a⃗=5ı^a⃗=5îa⃗=5ı^, b⃗=12k^b⃗=12k̂b⃗=12k^ and c⃗=a⃗×b⃗c⃗=a⃗\times b⃗c⃗=a⃗×b⃗ , find c⃗c⃗c⃗.

c ⃗ = − 60 ȷ ^ c⃗=-60ĵ c ⃗ = − 60 ȷ ^

If vectors a ⃗ = 5 ı ^ a⃗=5î a ⃗ = 5 ı ^ , b ⃗ = 12 k ^ b⃗=12k̂ b ⃗ = 12 k ^ and c ⃗ = a ⃗ × b ⃗ c⃗=a⃗\times b⃗ c ⃗ = a ⃗ × b ⃗ , find c ⃗ c⃗ c ⃗ .

Let's see how we can solve this problem. So in this problem, we're told if vector 'a' is equal to 5 'i', vector 'b' is equal to 12 ks, and vector c is the cross product of a and b, find vector c. So this is a situation where we need to do a cross product, but notice that we're not actually given a nice table to help us in this problem. What we have to do is construct this table ourselves, and it's important to have this skill where you know how to construct these tables when dealing with a cross product because it's not always going to be given to you. So let's go ahead and give this a try. So we need to find the cross product of a and b, respectively. We're trying to find a cross b. Now what I recall from the table that we've been dealing with is we have some kind of matrix that goes right here, and this matrix has I, j, and k components. Now what I also recall from this table is that the I and j components get repeated outside here, because what we need to do is repeat the columns that we get for I and j outside the matrix. Now let's actually try poking in the vectors. So you can see that vector a is 5 I, so that means we're going to have 5 right in the I column there, but I noticed that we don't have a j or a k, so these are just both going to be 0. Now you can see for b, we have 12 k. So you can put a k or a 12 here in the k, but we have zeros for I and j. Now what we need to do is repeat the I and the j columns. So we're going to have the I column, which is 50, then we're going to have the j column, which is 0. Now we have our matrix and our table set up, so we can do this cross product. Now the cross product is going to tell us the components of vector c. So we're gonna find the x component of c, the y component of of z of c, excuse me, and then the z component of c. Now what I can do is use this cross product for I. So I can go through I and I could cross here, and then cross up there. So this is gonna give us the x, and for the x we're going to have a 0 times 12, which is 0, minus 0 times 0, which is 0. So our x component is just going to be 0 I. Now what I can do is do the cross product for the y component. So if I go to j, and I cross through here, and cross up there, well we're going to have 0 times 0 which is 0, minus 12 times 5 which is 60. So we're going to have 0 minus 60 which is negative 60 j. Now lastly, we'll go to the z component which I can start here at k, cross through here, and cross up there. Well I can see that we'll have 5 times 0 which is 0, minus 0 times 0 which is 0, and 0 minus 0 will give us 0 k. So the vector we end up getting, is we end up getting that vector c is equal to just negative 60 j, because we've got both zeros for I and k, so we don't need to write those. Now if you wanted to write this in bracket notation, it would look like this. It would look like 0, and then we have the j component, which is negative 60, and we have the k component, which is 0. So these right here would be the correct answers or the solutions to this problem, and I want you to notice something. When we started with a vector that had only an I and a k component, we got a vector with a j component, and that's because the cross product gives you a perpendicular vector to the other 2. So it makes sense that we got a j component, and that's something we honestly could have predicted based on what we were given. So hope you found this video helpful. Thanks for watching.