𝐴𝐵𝐶𝐷 is a square of side four, and 𝐮 is a unit vector perpendicular to the square’s plane. Find 𝐀𝐃 cross 𝐁𝐂.
Looking at our square, we see that corners 𝐴 and 𝐷 are in these spots and 𝐵 and 𝐶 are here. Our question asks us to define vectors, one from point 𝐴 to point 𝐷 and the other from point 𝐵 to point 𝐶, and then cross them. Vector 𝐀𝐃 looks like this and 𝐁𝐂 like this. Since these vectors are on the sides of this square, we know that both of them have a magnitude of four. They both also point in the same direction in the plane of our screen.
Now, in general, if we have two vectors, we’ll call them 𝐀 and 𝐁, and we cross them, then that’s equal to the magnitude of one vector times the magnitude of the other times the sine of the angle between them. Here we’ve called that angle 𝜃, all in the direction of a vector that’s perpendicular both to vector 𝐀 and vector 𝐁.
Applying this relation to our scenario, on the right-hand side, we again use 𝜃 to specify the angle between our two vectors 𝐀𝐃 and 𝐁𝐂. This vector 𝐮 we’re told is a unit vector that’s perpendicular to the square’s plane. That is, it either goes into or out of the screen. Now, we know both the magnitude of 𝐀𝐃 and the magnitude of 𝐁𝐂. They’re both four. But then let’s think about the sine of the angle between these vectors. Since both vectors point in exactly the same direction, the angle between them is zero degrees.
We can write then that this cross product equals four times four times the sin of zero degrees in the 𝐮-direction. But then, the sin of zero degrees is itself equal to zero. And that means our entire expression, this cross product, is zero. Whenever two vectors are parallel, as they are in this case, or antiparallel, then the cross product between them must be zero. That’s because, for parallel vectors, the sin of zero degrees is zero. And for antiparallel vectors, the sin of 180 degrees is zero. 𝐀𝐃 cross 𝐁𝐂 then equals zero.