evaluate the dot product of (2 ,4) and (1, -2)
Yay! Dot products are easy! :)
All you have to do is multiply the first number with the first number, and the second number with the second number. Then, add together. It sounds complicated in words, but when you do it with the numbers it's simple. For example, if I want to find the dot product of (1,3) and (4,5) I would just do: (1)(4) + (3)(5) = 4 + 15 = 19.
thats it
For your question, we have (2,4) and (1,-2). To find the dot product we just do: (2)(1) + (4)(-2) = 2 - 8 = -6
And that's it, the dot product of (2,4) and (1,-2) is -6.
Does that make sense? Any questions? :)
yea
Cool!
The "cross" product is a little more complicated. But the "dot" product is easy.
what about a projection
Projection uses the dot product and the length of the coordinate (or vector). An extra step or two, but the same "math difficulty"
so (2,5) onto (1,3) would be sovled how
OK, I like drawing a picture to get an idea of what is even happening.
lol, after drawing it I change my mind, the picture confused me more. I'll just use the formula. :)
it has the answers of 0.5(1,3) 1.7(1,3) 1.7(2,5) 0.5(1,3)
what is th eforumla
Alright, so the formula is: let a = (x1,y1) and b = (x2,y2) projection of b onto a = (a dot b)/(a dot a) x a
where the "x a" means normal multiplication with a
All you need to worry about is finding (a dot b) and (a dot a)
Let's call a = (1,3) and b (2,5). Since the formula is project b onto a. We want to project (2,5) onto (1,3).
a dot b = (1,3) dot (2,5) = (1)(2) + (3)(5) = a dot a = (1,3) dot (1,3) = (1)(1) + (3)(3) =
a dot b = 17 a dot a = 10 any questions so far? Sometimes the notation can freak you out. :)
no i am good
If you've got all that, then the formula just says you divide (a dot b)/(a dot a) = 17/10 = 1.7 and then multiply that by a. The final answer would then be 1.7a = 1.7 (1,3) remember that we are calling a = (1,3)
Alright, so a lot more than "one or two extra steps" (I lied), but not THAT big of a math deal... lol
i think i got it
yea i got it
Cool cool, if you have any other questions let us know. Good luck!
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