## anonymous 4 years ago the symbol$A _{b}$ stands for the projection of vector A onto vector B. In other words, $A _{b}$ represents the component of A that is parallel to B. Derive an expression for $A _{b}$ in terms of the vectors A and B.

1. amistre64

$\frac{a.b}{|a|^2}a$

2. anonymous

wow u did that really quickly

3. amistre64

just got out of calc3 that just went over it :)

4. anonymous

in the numerator...is that a times b?

5. amistre64

a dot b

6. anonymous

a dot b divided by magnitude of a squared all multiplied by a?

7. amistre64

yes, the left side is a scalar; and the right side is vector a scaled to that length

8. anonymous

would it be a bother to run me through how u got to this expression?

9. amistre64

|dw:1327527260794:dw|

10. amistre64

|b| cos(t) = the length of Ab, if we use your notation for proj{a} b

11. amistre64

$|b|cos(t) =|b| \frac{a.b}{|a||b|}=\frac{a.b}{|a|}$ good so far?

12. anonymous

yea i get that

13. anonymous

i'm with u

14. anonymous

i'm also trying to decode your picture :)

15. anonymous

is that the x coordinate down there at the end?

16. anonymous

the long vector

17. amistre64

now, we need to scale that to a unit vector of a; since a is |a| long, lets divide it by |a| to get it to a length of 1

18. amistre64

$\frac{a.b}{|a|}\ \frac{a}{|a|}=; unit\ a, scaled\ by\ needed\ length$

19. anonymous

okay

20. amistre64

and since |a| |a| = |a|^2 i just condensed it alittle bit

21. anonymous

simple enough?

22. amistre64

|dw:1327527605851:dw|

23. anonymous

can u redraw that picture from above?

24. anonymous

oh okay now i see

25. amistre64

lol, i cant draw on this thing very well period :)

26. anonymous

okay....so that's it for the equation now?

27. amistre64

to find the the vector that is the proj of b onto a; yes; reduce a to ints unit equivalent; and scale it up by the magnitude of b*cos(t)

28. anonymous

and what u first initally gave me is that correct?

29. amistre64

yes, seeing how that is what we ended up with in the end i would say that its correct :)

30. anonymous

u are an absolute life saver

31. amistre64

thnx, good luck :)