# need some help from fellow math geeks

Discussion in 'Members' Lounge' started by GSRCRXsi, Jan 23, 2004.

1. ### GSRCRXsiSuper ModeratorModeratorVIP

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ok i need a little help for my multivariable calc class. we have this assignment, and i cant figure out this one problem for the life of me. we are doing vectors.

the problem:

write c in the form ra + sb, where r and s are scalars.

a = 3i + 2j, b = 8i + 5j, c = 7i + 9j

now basically i have to set
c = ra + sb
or
7i + 9j = r(3i + 2j) + s(8i + 5j)
and find the actual value of r and s but i just dont know how to go about it, ive been trying to do it with trial and error but nothing has worked so far. is there a special way i should go about this or manipulate it to make it work?. anyone have any tips?

2. ### TurboMirageYEEAAAHHHVIP

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"damnit jim im a doctor not a math teacher!!"

3. ### GSRCRXsiSuper ModeratorModeratorVIP

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ok this is what i have so far....

i distributed the r and s: (r3i + r2j) + (s8i + s5j)

combine like terms: (r3i + s8i) + (r2j + s5j)

pull out the coefficients: (3r + 8s)i + (2r + 5s)j

now (3r + 8s) has to = 7, because i need 7i for the i component
and
(2r + 5s) has to = 9 becasue i need 9j for the j component

now i did some algebra:

7 = 3r + 8s 7 = 2r + 5s - 2

3r + 8s = 2r + 5s -2
3r - 2r = 5s - 8s - 2
r= -3s - 2

plug the r substitution back into the equation:

3(-3s - 2) + 8s = 2(-3s - 2) + 5s - 2
-9s -6 +8s = -6s -4 + 5s -2
-s - 6 = -s - 6
s=s

now i just proved that s=s, WTF??? thats not what i need, i need to find the numerical answer. what am i doing wrong?

4. ### TommyTheCatGonzo Scientist

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using, find c = ra +sb
a = 3i + 2j, b = 8i + 5j, c = 7i + 9j

By substitution, we find that c = 7i + 9j = r(3i + 2j) + s(8i + 5j)

Now, let us simplify by concentrating on only one of the vectors.

7i = r(3i) + s(8i) and 9j = r(2j) + s(5j), ignore the vector components now:

7 = 3r + 8s
9 = 2r + 5s

Let's use the j vector comp. equa. and solve for s.

9 = 2r + 5s
9 - 2r = 5s
(9 - 2r) / 5 = s

Now put this s value defined by r into the other equa.

7 = 3r + 8(9 - 2r) / 5

By out algebra, we simplify and get:

r = 37

Plus this value into either equa. to find s:

7 = 3r + 8s
7 = 3(37) + 8(s)
s = -13

Any questions?

5. ### TommyTheCatGonzo Scientist

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When you did this:

7 = 3r + 8s 7 = 2r + 5s - 2

3r + 8s = 2r + 5s -2
3r - 2r = 5s - 8s - 2
r= -3s - 2

You were ignoring the i and j parts. They are really still there. So you did this:

7i = 3ir + 8is 7j = 2jr + 5js - 2

and 7i =/= 7j
you cant take that step, you have to solve within the respective systems.

6. ### Def_JamSenior Member

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I knew I should have sleep less in highschool and went to collage afterward.

7. ### Gen2TegSenior Member

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8. ### randerson165Senior Member

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Thats some crazy math there. Makes me wanna go back to school and learn.

9. ### TommyTheCatGonzo Scientist

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It's really not that hard when you learn how to ignore some things. I'm only starting CalcII with vector stuff (all the i and j junk) and we're not doing multivariables (CalcIII). This is pure algebra work here, separating variables from numbers/scalars and finding values for the variables.

But, GSRCRXsi, just wondering if you got it.

10. ### GSRCRXsiSuper ModeratorModeratorVIP

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yea i got it tommy, thanks. i was trying to set the 2 equations equal to each other when i shoulda done each one seperate, i always mess up little details like that, lol. too bad i turned in the assignment before i got your explaination. i just left it as c = 7i + 9j = r(3i + 2j) + s(8i + 5j), lol. whatever, its not that important, i jus wanted to make sure I understood what was happening. i had calcII last semester and we touched on vectors at the very end. the vector stuff is somewhat review, and then we are going on from that. we are getting into dot products and all kinds of fun shit now . and i will need ANOTHER semester of math as well calc IV baby, and im only a damn freshman, lol. all the people in my calc 3 class are upper classmen, except me and 1 other kid.

now i did some of this stuff in highschool too, jus wondering tommy, what year are you in in college (if in college) and whats your major?

thanks again for the help tommy, and to everyone else, dont you wish you knew what was going on?

11. ### CRX-YEMSuper ModeratorModeratorVIP

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Differential Equations (aka Calc IV), I've got to take that class as well.
Calc III is way easier than Calc II

12. ### GSRCRXsiSuper ModeratorModeratorVIP

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yea calc 2 has so much crammed into it

13. ### B16Super ModeratorVIP

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i went through calc II, barely made it

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:sleep:

15. ### GSRCRXsiSuper ModeratorModeratorVIP

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come on tony, its that that bad

16. ### MaaseyRacerSenior Member

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I am starting calc one on monday and you guys now have me scared as I understood nothing of the above.

17. ### GSRCRXsiSuper ModeratorModeratorVIP

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dont worry dude, thats calc 2/3 shit, calc 1 is stupid easy. just limits taking derivatives and shit, its easy, i swear, lol.

18. ### CRX-YEMSuper ModeratorModeratorVIP

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yeah if you can multiply and divide
you can easily do calc 1 it's cake

19. ### jiahanhaoSenior Member

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i miss calc...

calc >>>> combinatorics and optimization

i could sing an ode to O.D.E's...

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