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Spatial Vectors 3

Vectors… in… Space….!



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Welcome to a mini-math series about spatial vectors. In part 1, we’ve introduced vectors and put them in space. Now in part 2 we saw some elementary things we can do with them. Now we’ll look at vectors products!


How cool does that sound?


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Seriously now, it’s vector products time

The first important vector product is called the scalar, or simply dot product. It’s defined as: a \cdot b = ||a|| ||b|| \cos \theta, where ||a|| is the magnitude or length of a and \theta is the angle between the tails of a and b.

The dot product returns a scalar result, not a vector, so you may get 5 or -22 or 2532.253588836.

There’s another fun and easy way to find the dot product, which works better in 3D since we don’t really have an angle measure. This formula work like this: for vectors a=(x_1,y_1,z_1), b=(x_2,y_2,z_1), we have that a \cdot b = x_1x_2+y_1y_2+z_1z_2. This would work similarly for more variables in higher dimensions.

Clearly, since multiplication and addition are independent of direction, a \cdot b=b \cdot a.

The whole point of the dot products is that it returns the product of the magnitudes of two vector, but only for the components that point in the same direction. It’s a useful thing to be able to do.

An important application of this is work, in physics. Work deals with only the magnitude of force that acts in the direction of the motion, so if we have this box pushed with force F over distance d:

Then we use the dot product of vectors F and d to find the work done. What a nice application!

Now, there’s another product we’ve got to talk about. This one is called the cross product. It’s only defined in 3D, which is important to remember. What the cross product does is return a vector that is perpendicular to both original vectors (which would only be possible and meaningful in 3D).

It is defined as a \times b=||a|| ||b|| \sin \theta \hat n, again with ||a|| representing the magnitude and \theta the angle between the tails of the vectors. \hat n represents a unit vector that is perpendicular to the original 2. Basically, the first portions (magnitudes and signs) identify the appropriate size of the cross product, while \hat n provides the direction.

Note that this time, a \times b \not= b \times a. This is because there are two perpendicular vectors of equal magnitude and opposite directions. This means that a \times b = - (b \times a). Now, the direction of the perpendicular unit vector is determined by the right hand rule:

Again, there is a simpler formula, although it’s a bit harder to memorize than the dot product. It goes like this: for a=(x_1,y_1,z_1), b=(x_2,y_2,z_2) we’ve got a \times b= (y_1z_2-z_1y_2, z_1x_2-x_1z_2, x_1y_2-y_1x_2).

Note that if the 2 original vectors are parallel, having the same direction, then the angle between them would force the cross product to be zero- which makes perfect sense, it’s not very possible to find just one specific vector perpendicular to a certain direction.

Another cool application is figuring out the volume of a parallelepiped (a box that doesn’t necessarily have all sides as rectangles, they’re just parallelograms) with 3 vectors as sides. The formula is called the triple scalar product: V= a \cdot (b \times c).

And hey, this is pretty much it. Basic vectors don’t have much more to them than this! Now you should make sure to have some practice and enjoy a soothing, relaxing vectors test. How fun would that be?


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Vectors… in… Space….!

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