Question

Here's what I have done:
(5,3)x(7,-2)/|(7,-2)| = 29/sqrt53 = 3.98

I thought I was done, but it says there is more to the problem. What do I do from here???

Answer 1

\[proj_a b=\frac{\vec a.\vec b}{|a|^2}\ \vec a\]

Answer 2

@rebbylack look carefully. and understand what amistre64 says. He is great in mathematics. good luck.

Answer 3

(5,3)
(7,-2) : |..| = sqrt(53)
--------
35-6 = 29

\[\frac{29}{53}<7,-2>\]

Answer 4

in effect, we are taking a unit vector in the direction of a and scaling it by the value |b| cos(theta)

Answer 5

this gives us then the length and the direction we want

Answer 6

\[\cos\theta =\frac{a.b}{|a||b|}\]
\[|b|\cos\theta =|b|\frac{a.b}{|a||b|}=\frac{a.b}{|a|}\]
scale this in the direction of the unit vector of a

\[|b|\cos\theta \frac{a}{|a|}=\frac{a.b}{|a|}\frac{a}{|a|}\to\ \frac{a.b}{|a|^2}\ a \]

Answer 7

I think I'm beginning to understand... but I found the solution in my book and it says to use \[\frac{3.98}{\sqrt{53}} <5,3>\] ??? Does this way and your way both work?

Answer 8

Oh wait...I see it now!!! they both equal the same thing. amistre64, you're a genius!! thank you sooo much :)

Answer 9

you stated you want 5,3 ONTO 7,-2

so it important that your direction of the projection is in line with 7,-2 otherwise its not a projection ONTO 7,-2

Answer 10

your last post has this thing running with <5,3> instead of <7,-2>