Video: Finding the Dot and Cross Products between Vectors

Alex Cutbill

If 𝐴 = 〈−3, 0, −2〉, 𝐵 = 〈−1, −3, 3〉, and 𝐶 = 〈2, −2, −1〉, find (𝐴 + 𝐶) ⋅ [(𝐴 × 𝐵) × (𝐵 × 𝐶)].

06:19

Video Transcript

If 𝐴 is the vector with components negative three, zero, negative two; 𝐵 is the vector with components negative one, negative three, three; and 𝐶 is the vector with components two, negative two, negative one, find 𝐴 plus 𝐶 dot 𝐴 cross 𝐵 cross 𝐵 cross 𝐶.

There isn’t really a trick to this question. We just have to compute this. We use the order of operations to break this big problem up into smaller problems. For example, finding 𝐴 plus 𝐶, 𝐴 cross 𝐵, and 𝐵 cross 𝐶 before combining them. Let’s start with finding 𝐴 plus 𝐶. We simply add the components of 𝐴 and 𝐶 to get negative one, negative two, negative three. Now we move on to finding the cross product of 𝐴 and 𝐵. We represent this cross product as a determinant where the entries in the first row are the unit vectors 𝑖, 𝑗, and 𝑘 which point in the 𝑥-, 𝑦-, and 𝑧-directions, respectively. In the second row, we put the components of 𝐴. So that’s negative three, zero, negative two. And in the third row, we put the components of 𝐵. So that’s negative one, negative three, three.

Now we just need to evaluate this determinant in the normal way. We expand along the first row getting a term for each of the entries 𝑖, 𝑗, and 𝑘. We now just need to evaluate each of the two-by-two determinants. And we can write this in component form like so. Let’s clear the working away to make room. Now we work on finding 𝐵 cross 𝐶. It’s the same process. We fill the three-by-three determinant with the unit vectors 𝑖, 𝑗, and 𝑘 and the components of 𝐵 and 𝐶; expand along the first row; and evaluate the two-by-two determinants before expressing this in component form. We now have all the ingredients prepared.

Now that we have the values of 𝐴 cross 𝐵 and 𝐵 cross 𝐶, we can evaluate the cross product of cross products, 𝐴 cross 𝐵 cross 𝐵 cross 𝐶. We write the cross product as a three-by-three determinant, we expand along the first row, and evaluate the two-by-two determinants before writing the vector in component form. Now we can compute the dot product. We have the components of the vector 𝐴 plus 𝐶 and the components of our cross product of cross products. We multiply corresponding components and add the products up.

And evaluating this, we find that our answer is 86.

I made a claim at the beginning of the video that there wasn’t a trick to this question. We just had to follow the order of operations carefully and compute. The rest of this video will be about higher-level concepts that we could’ve used to save us a bit of computation, but not much. But before we get on to that, I’d like to point out something that would’ve required less computation but also, unfortunately, gives the wrong answer. So on balance, it was probably worth doing the work.

You might have been tempted to regroup the factors in the cross product of cross products. The factors remain in the same order. But we’d move some parentheses around. The reason that it’s tempting to do this is that the cross product of a vector with itself is just the zero vector. So 𝐵 cross 𝐵 certainly is the zero vector. And any vector crossed with the zero vector is also the zero vector. So 𝐴 cross 𝐵 cross 𝐵 is also the zero vector. And again, the zero vector crossed with any vector is the zero vector. So the whole thing is surely just the zero vector. And when you dot 𝐴 plus 𝐶 with this zero vector, of course you’re going to get zero. Unfortunately, as we’ve discovered, the correct answer is actually 86 and not zero. And along the way, we showed that the cross product of cross products isn’t the zero vector.

So what went wrong? It is true that 𝐴 cross 𝐵 cross 𝐵, or crossed with 𝐶, is the zero vector. It was the first step we made when we rearranged parentheses that was invalid. Unlike real products involving the multiplication of real numbers and even matrix products involving the multiplication of matrices, the cross product of vectors is not associative. This means that we can’t just move parentheses around as we like in order to regroup terms. We end up changing the value as we’ve seen.

While the cross product of cross products is not zero, you might have noticed that it’s a multiple of the vector 𝐵. If we think about it, this isn’t too surprising. The cross product returns a vector which is perpendicular to both inputs. We know, for example, that 𝐴 cross 𝐵 is perpendicular to both 𝐴 and 𝐵. And 𝐵 cross 𝐶 is perpendicular to both 𝐵 and 𝐶. So 𝐵 and all its multiples are perpendicular to both 𝐴 cross 𝐵 and 𝐵 cross 𝐶. Our cross product of cross products is also perpendicular to both 𝐴 cross 𝐵 and 𝐵 cross 𝐶. And so, it’s no surprise that it’s a multiple of 𝐵. The question is, which multiple of 𝐵 is it? It turns out that you can prove for any vectors 𝐴, 𝐵, and 𝐶 that our cross product of cross products would be 𝐴 cross 𝐵 dot 𝐶 times 𝐵. And you can check that in our example, 𝐴 cross 𝐵 dot 𝐶 is negative 43.

Proving this identity takes quite a bit of work. And even applying it requires quite a bit of computation, although not as much as without it. To conclude, for this problem it was probably best just to carefully compute. But if you’re still watching, I hope you enjoyed this preview of higher-level vectors stuff.

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