If vectors v⃗=⟨3,1,0⟩v ⃗=⟨3,1,0⟩v⃗=⟨3,1,0⟩, u⃗=⟨0,−2,0⟩u ⃗=⟨0,-2,0⟩u⃗=⟨0,−2,0⟩, and w⃗=v⃗×u⃗w ⃗=v ⃗×u ⃗w⃗=v⃗×u⃗, find w⃗⃗w⃗⃗w⃗⃗.

w ⃗ = ⟨ 0 , 0 , − 6 ⟩ w⃗=\langle0,0,-6\rangle w ⃗ = ⟨ 0 , 0 , − 6 ⟩

If vectors v ⃗ = ⟨ 3 , 1 , 0 ⟩ v ⃗=⟨3,1,0⟩ v ⃗ = ⟨ 3 , 1 , 0 ⟩ , u ⃗ = ⟨ 0 , − 2 , 0 ⟩ u ⃗=⟨0,-2,0⟩ u ⃗ = ⟨ 0 , − 2 , 0 ⟩ , and w ⃗ = v ⃗ × u ⃗ w ⃗=v ⃗×u ⃗ w ⃗ = v ⃗ × u ⃗ , find w ⃗⃗ w⃗⃗ w ⃗⃗ .

Let's see how we can solve this problem. Here we're told if vector v is equal to 3, comma, 1, comma, 0, and vector u is equal to 0, comma, negative 2, comma 0, and w is the cross product of v and u, find w. So this is a problem where we need to calculate the cross product, and in this problem, we're actually given this table that we've talked about that allows you to find the cross product between vectors. So let's go ahead and begin. Now for vector v, we're going to put the I, j, and k components that we see there in for v. So we can see that I is 3, j is 1, and k is 0. So we're going to put 310 for vector v. Now for vector u, that's going to be the I j k components of that vector. So we can see based on what we're given it's 0, negative 2, 0 that's going to be vector u. Now what I can also see here is that I need to repeat the I and j column. So I can see that the I column is 3 and then 0, because we have the vector v and the vector u, and then I can see for this column for the j's, we have 1 and negative 2. So now we've gone ahead and plugged in all the numbers, which means we need to do the cross product operation from here. Now what the cross product operation is going to tell you is what the x component is, the y component is, and the z component is, or basically the I, the j, and the k. So let's go ahead and figure this out. Now I'll start here with I, and what I'm going to do is cross through here, and then cross up there. So we're going to have 1 times 0 which is 0, and then we'll have negative 2 times 0 which is 0, and then that's going to be 0 minus 0. Now let's go to the y component which is j. If I go here through j and cross there and there, we'll have 0 times 0 which is 0 minus 0 times 3 which is 0. Now what I'll do is I'll last go to the k component, I'll cross through here and I'll cross up there. So we'll have 3 times negative 2, and 3 times negative 2 is negative 6, and that's going to be minus 0 times 1, which is 0. So this is what we end up having for the I component, for the j component, and for the k component of our vector, because recall that the cross product gives you another vector rather than a scalar. So what we can do is complete these operations. So we'll have 0 minus 0 which is 0 I, we'll have 0 minus 0 which is 0 j, and we'll have negative 6 minus 0 which is negative 6 k. So what our vector ends up becoming if we do this in the bracket notation, is we'll have that vector w is equal to 0, comma, 0, comma, negative 6, and this right here is the vector we end up getting, and that is the solution to this problem. Now I do want you to notice something interesting about the result that we got. Notice how the first vector we had had an I and j component, The second vector we had had just a j component, but the third vector we got the vector that came out with this cross product was a k component, and it's just a k component. Well, this actually makes sense, but if you think about it, the cross product gives you a vector that's perpendicular to the other 2. So since both of these were only existing in either I or j, it makes sense that we got a vector that only had a k component. Now don't worry, that's not something you need to memorize per se, but it's just something that's interesting about the result that we got, and it's something you should recognize when dealing with cross products. So hope you found this video helpful. Thanks for watching.

If vectors v ⃗ = ⟨ 3 , 1 , 0 ⟩ v ⃗=⟨3,1,0⟩ v ⃗ = ⟨ 3 , 1 , 0 ⟩ , u ⃗ = ⟨ 0 , − 2 , 0 ⟩ u ⃗=⟨0,-2,0⟩ u ⃗ = ⟨ 0 , − 2 , 0 ⟩ , and w ⃗ = v ⃗ × u ⃗ w ⃗=v ⃗×u ⃗ w ⃗ = v ⃗ × u ⃗ , find w ⃗⃗ w⃗⃗ w ⃗⃗ .

w ⃗ = ⟨ 0 , 0 , − 6 ⟩ w⃗=\langle0,0,-6\rangle w ⃗ = ⟨ 0 , 0 , − 6 ⟩

w ⃗ = ⟨ 0 , − 2 , 0 ⟩ w⃗=\langle0,-2,0\rangle w ⃗ = ⟨ 0 , − 2 , 0 ⟩

w ⃗ = ⟨ 0 , 0 , 6 ⟩ w⃗=\langle0,0,6\rangle w ⃗ = ⟨ 0 , 0 , 6 ⟩

w ⃗ = ⟨ 0 , 0 , − 2 ⟩ w⃗=\langle0,0,-2\rangle w ⃗ = ⟨ 0 , 0 , − 2 ⟩

8. Vectors

Cross Product

Trigonometry

8. Vectors

Cross Product

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Multiple Choice

If vectors $v ⃗=⟨3,1,0⟩$ , $u ⃗=⟨0,-2,0⟩$ , and $w ⃗=v ⃗×u ⃗$ , find $w⃗⃗$ .

$w⃗=\langle0,0,-6\rangle$

$w⃗=\langle0,-2,0\rangle$

$w⃗=\langle0,0,6\rangle$

$w⃗=\langle0,0,-2\rangle$

Verified step by step guidance

To find the cross product $ \mathbf{w} = \mathbf{v} \times \mathbf{u} $, we use the determinant of a 3x3 matrix. The first row of the matrix consists of the unit vectors $ \hat{i}, \hat{j}, \hat{k} $.

The second row of the matrix consists of the components of vector $ \mathbf{v} = \langle 3, 1, 0 \rangle $, so it will be $ 3, 1, 0 $.

The third row of the matrix consists of the components of vector $ \mathbf{u} = \langle 0, -2, 0 \rangle $, so it will be $ 0, -2, 0 $.

Calculate the determinant of the matrix: $ \mathbf{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 0 & -2 & 0 \end{vmatrix} $.

Expand the determinant: $ \mathbf{w} = \hat{i}(1 \cdot 0 - 0 \cdot (-2)) - \hat{j}(3 \cdot 0 - 0 \cdot 0) + \hat{k}(3 \cdot (-2) - 1 \cdot 0) $. Simplify each term to find the components of $ \mathbf{w} $.

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