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Question Video: Calculating the Vector Product of Two Vectors That Lie in the π‘₯𝑦-Plane Given Their Components Physics

Consider two vectors 𝐂 = 15𝐒 + 7𝐣 and 𝐃 = 4𝐒 + 9𝐣. Calculate 𝐂 Γ— 𝐃. Calculate 𝐃 Γ— 𝐂.

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Video Transcript

Consider two vectors 𝐂 equals 15𝐒 plus seven 𝐣 and 𝐃 equals four 𝐒 plus nine 𝐣. Calculate 𝐂 cross 𝐃. Calculate 𝐃 cross 𝐂.

We are tasked with calculating two cross products, also called vector products, using the vectors 𝐂 and 𝐃 defined at the top. Whenever we need to perform multiple calculations, it’s worth taking a moment to see if those calculations are related. In this particular case, the calculations are directly related. It is a property of the vector product that changing the order of the two vectors only changes the sign of the final answer. Algebraically, we write 𝐂 cross 𝐃 is equal to negative 𝐃 cross 𝐂. So, we actually only need to compute one of these cross products, either 𝐂 cross 𝐃 or 𝐃 cross 𝐂, because we can then find the other one immediately by flipping the sign.

Alright, so instead of two vector products, we can actually answer both parts of this question after only computing one vector product. So, let’s see how to form the vector product of two vectors. We can write a simple formula for this vector product by noting that 𝐂 and 𝐃 are both written in terms of the 𝐒 and 𝐣 unit vectors, and 𝐒 and 𝐣 point, respectively, in the positive π‘₯- and positive 𝑦-directions of the Cartesian set of axes. To understand why this is important, we have drawn a set of three-dimensional Cartesian axes, where π‘₯ increases to the bottom left, 𝑦 increases to the right, and 𝑧 increases upward. Note, of course, that this is only a two-dimensional picture of what is in reality a three-dimensional object.

Anyway, the vector 𝐒 is defined as pointing in the positive π‘₯-direction and having a magnitude of one. The vector 𝐣 is defined as pointing in the positive 𝑦-direction and having a magnitude of one. And as we can see, this leaves space for a third unit vector pointing in the positive 𝑧-direction also with a magnitude of one, and we call this vector 𝐀. Now, the vector product of two nonparallel vectors is always perpendicular to the plane that contains those two vectors. Since the vectors 𝐂 and 𝐃 are both expressed in terms of the 𝐒 and 𝐣 unit vectors only, they both exist in the π‘₯𝑦-plane. And so their vector product must be perpendicular to the π‘₯𝑦-plane, that is, in the positive 𝐀-direction or possibly in the negative 𝐀-direction.

Now, remember that 𝐂 cross 𝐃 and 𝐃 cross 𝐂 only differ by a negative sign. So, one of these will be in the positive 𝐀-direction and one of these will be in the negative 𝐀-direction. Now, because these vector products will be in the 𝐀- or negative 𝐀-directions, we can write the simple formula 𝐂 cross 𝐃 is the π‘₯-component of 𝐂 times the 𝑦-component of 𝐃 minus the 𝑦-component of 𝐂 times the π‘₯-component of 𝐃 in the 𝐀-direction. If the quantity in the parentheses is positive, the direction will be positive 𝐀. And if the quantity in the parentheses is negative, the direction will be negative 𝐀.

We can also see from this formula why changing the order of the vectors changes the sign of the vector product. When we exchange 𝐂 and 𝐃 on the left side of this equation, we also need to exchange the two terms in the parentheses. But exchanging those two terms is equivalent to multiplying the whole thing by negative one. We should also mention that when we write a vector as a sum in terms of 𝐒 and 𝐣 unit vectors, the π‘₯-component is whatever number is multiplying the 𝐒 unit vector, and the 𝑦-component is whatever number is multiplying the 𝐣 unit vector. This is perfectly logical. As we can see from the bottom-left diagram, 𝐒 is along the π‘₯-axis and 𝐣 is along the 𝑦-axis.

Now, all we need to do is substitute numbers and calculate. Starting with 𝐂 cross 𝐃, we have 15 times nine, which is 𝐢 π‘₯ times 𝐷 𝑦, minus seven times four, which is 𝐢 𝑦 times 𝐷 π‘₯, in the 𝐀-direction. 15 times nine is 135, and seven times four is 28. So, we have 135 minus 28, which is 107, in the 𝐀-direction. This gives us 𝐂 cross 𝐃, and all we need to do to find 𝐃 cross 𝐂 is multiply by negative one. So, 𝐂 cross 𝐃 is 107𝐀, and 𝐃 cross 𝐂 is negative 107𝐀. Or equivalently, 𝐂 cross 𝐃 is 107 in the 𝐀-direction, and 𝐃 cross 𝐂 is 107 in the negative 𝐀-direction.

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