Question Video: Calculating the Vector Product of Two Vectors Shown on a 3D Grid Physics

The diagram shows two vectors, 𝐀 and 𝐁, in three-dimensional space. Both vectors lie in the π‘₯𝑦-plane. Each of the squares on the grid has a side length of 1. Calculate 𝐀 Γ— 𝐁.

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Video Transcript

The diagram shows two vectors, 𝐀 and 𝐁, in three-dimensional space. Both vectors lie in the π‘₯𝑦-plane. Each of the squares on the grid has a side length of one. Calculate 𝐀 cross 𝐁.

This question is asking us about vector products. Specifically, we are asked to work out the vector product 𝐀 cross 𝐁, where the vectors 𝐀 and 𝐁 are given to us in the form of arrows drawn on a diagram, and we are told that both vectors lie in the π‘₯𝑦-plane.

Let’s begin by recalling the definition of the vector product of two vectors. We’ll consider two general vectors, 𝐂 and 𝐃, and suppose that both of them lie in the π‘₯𝑦-plane. Then we can write these vectors in component form as an π‘₯-component labeled with a subscript π‘₯ multiplied by 𝐒 hat plus a 𝑦-component labeled with a subscript 𝑦 multiplied by 𝐣 hat. Remember that 𝐒 hat is the unit vector in the π‘₯-direction and 𝐣 hat is the unit vector in the 𝑦-direction. Then the vector product 𝐂 cross 𝐃 is given by the π‘₯-component of 𝐂 multiplied by the 𝑦-component of 𝐃 minus the 𝑦-component of 𝐂 multiplied by the π‘₯-component of 𝐃. And this is all multiplied by 𝐀 hat, which is the unit vector in the 𝑧-direction.

This general expression for the vector product is telling us that if we want to calculate 𝐀 cross 𝐁, then we’re going to need to work out the π‘₯- and 𝑦-components of the vectors 𝐀 and 𝐁. The vectors 𝐀 and 𝐁 are both shown in the diagram that’s given to us in the question. And the question also tells us that each of the squares on the grid in this diagram has a side length of one. This means that in order to find the π‘₯- and 𝑦-components of our vectors 𝐀 and 𝐁, all we need to do is count the number of squares that each vector extends in the π‘₯-direction and the 𝑦-direction.

Let’s begin by doing this for vector 𝐀. By tracing down from the tip of vector 𝐀 to the π‘₯-axis, we can see that vector 𝐀 extends one, two, three, four units in the negative π‘₯-direction. And by tracing across to the 𝑦-axis, we can see that 𝐀 extends one, two, three units in the positive 𝑦-direction. So we know that the π‘₯-component of 𝐀 is negative four and the 𝑦-component is positive three. In component form, we therefore have that the vector 𝐀 is equal to negative four 𝐒 hat plus three 𝐣 hat.

Now let’s do the same thing with vector 𝐁. If we trace down from the tip of vector 𝐁 until it meets the π‘₯-axis, then we can see that 𝐁 extends one, two, three, four squares into the positive π‘₯-direction. And if we trace across to the 𝑦-axis, we see that 𝐁 extends one, two, three, four, five squares into the positive 𝑦-direction. So the π‘₯-component of 𝐁 is positive four and the 𝑦-component is positive five. So writing vector 𝐁 in component form, we have that 𝐁 is equal to four 𝐒 hat plus five 𝐣 hat.

We now have each of our vectors 𝐀 and 𝐁 written in component form, which means we are now ready to calculate the vector product 𝐀 cross 𝐁. In our general expression for the vector product of two vectors, we see that the first term is given by the π‘₯-component of the first vector in the product multiplied by the 𝑦-component of the second vector in the product. In our case, the first vector in our product is 𝐀 and the second vector is 𝐁. So we need the π‘₯-component of vector 𝐀, which is negative four, multiplied by the 𝑦-component of vector 𝐁, which is five. We then subtract a second term from this first one.

This second term is the 𝑦-component of the first vector in the product multiplied by the π‘₯-component of the second vector in the product. So for us, that’s the 𝑦-component of vector 𝐀, which is three, multiplied by the π‘₯-component of vector 𝐁, which is four. And then this whole thing gets multiplied by the unit vector 𝐀 hat.

The final step is to evaluate this expression here. The first term, negative four multiplied by five, gives us negative 20. And the second term, three multiplied by four, gives us 12. When we calculate negative 20 minus 12, we get a value of negative 32. And so our answer to the question is that the vector product 𝐀 cross 𝐁 is equal to negative 32𝐀 hat.

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