Video: Finding the Cross Product of Two Parallel Vectors Represented in a Given Figure

๐ด๐ต๐ถ๐ท is a square of side 4, and ๐ฎ is a unit vector perpendicular to the squareโ€™s plane. Find ๐€๐ƒ ร— ๐๐‚.

01:59

Video Transcript

๐ด๐ต๐ถ๐ท is a square of side four, and ๐ฎ is a unit vector perpendicular to the squareโ€™s plane. Find ๐€๐ƒ cross ๐๐‚.

Looking at our square, we see that corners ๐ด and ๐ท are in these spots and ๐ต and ๐ถ are here. Our question asks us to define vectors, one from point ๐ด to point ๐ท and the other from point ๐ต to point ๐ถ, and then cross them. Vector ๐€๐ƒ looks like this and ๐๐‚ 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, weโ€™ll call them ๐€ and ๐, and we cross them, then thatโ€™s equal to the magnitude of one vector times the magnitude of the other times the sine of the angle between them. Here weโ€™ve called that angle ๐œƒ, all in the direction of a vector thatโ€™s perpendicular both to vector ๐€ and vector ๐.

Applying this relation to our scenario, on the right-hand side, we again use ๐œƒ to specify the angle between our two vectors ๐€๐ƒ and ๐๐‚. This vector ๐ฎ weโ€™re told is a unit vector thatโ€™s perpendicular to the squareโ€™s plane. That is, it either goes into or out of the screen. Now, we know both the magnitude of ๐€๐ƒ and the magnitude of ๐๐‚. Theyโ€™re both four. But then letโ€™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 ๐ฎ-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. Thatโ€™s because, for parallel vectors, the sin of zero degrees is zero. And for antiparallel vectors, the sin of 180 degrees is zero. ๐€๐ƒ cross ๐๐‚ then equals zero.

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