melimoo
knows how to rave
thanks a bunch guys!!
First year Math help (1 Viewer)
- Thread starter melimoo
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xuvu6520
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ooopse!!!SeDaTeD said:
heheheh.
btw. I dunno whose account is this. This guy was logged in from Mathematics Library.
LostAuzzie
Member
I'm stuck with question 4d.
Find the equation of the line that lies in plane p passes through point C and is perpendicular to the line l
I tried using the fact that the line is perpendicular to l and the normal to p, but that ended up giving me a plane
Anyone got any ideas, no answers just an idea of how to approach it
Any help would be much appreciated
Find the equation of the line that lies in plane p passes through point C and is perpendicular to the line l
I tried using the fact that the line is perpendicular to l and the normal to p, but that ended up giving me a plane
Anyone got any ideas, no answers just an idea of how to approach it
Any help would be much appreciated
melimoo
knows how to rave
i forget if i could do that one. i'll get back to you tomorrow after my notes have dried. yes, i did spill a whole full glass of water on all my math1002notes. fuckLostAuzzie said:
LoneShadow
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Post a copy of the assignment. Maths website is down or something. Or it's just me.LostAuzzie said:
LostAuzzie
Member
Yer it was down last night, should be up now though, if not heres the assignment
KeypadSDM
B4nn3d
It's the line in the plane P which is perpendicular to the projection of l onto P. Just work out the projection of l onto P, and then cross product its direction with the normal of P to find the direction of your required line.
Given the intersection coordinates you can find the equation of the line (i.e. you have the gradient (direction) and a point on the line)
Given the intersection coordinates you can find the equation of the line (i.e. you have the gradient (direction) and a point on the line)
LostAuzzie
Member
Thankyou for that, very much appreciatedKeypadSDM said:
And acmilan, only 2 of the questions are marked
melimoo
knows how to rave
ok here are my issues:
1c) find a unit vector perpecndicular to a and b+c
now i can find one perp. to a, but how do i find one perp to both?
2b) i figure i have to use the result from part a), and then use pythag. or sometihing...but how & where & why does a . b come into it?
2c) maybe would make more sense if i had b)??
also 4c) and d)
they make me cry.
any help is welcomed help and possibly repaid in sexual favours.
1c) find a unit vector perpecndicular to a and b+c
now i can find one perp. to a, but how do i find one perp to both?
2b) i figure i have to use the result from part a), and then use pythag. or sometihing...but how & where & why does a . b come into it?
2c) maybe would make more sense if i had b)??
also 4c) and d)
they make me cry.
any help is welcomed help and possibly repaid in sexual favours.
heart
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for 1c) it's just a x (b+c)... and then make the result into a unit vector
2b) Just say |AP|²= AP.AP and then a.b will come in the answer and do the same for |CP|²
c) this does make more sense if you do 2b
And i don't think i did 4c properly =P
2b) Just say |AP|²= AP.AP and then a.b will come in the answer and do the same for |CP|²
c) this does make more sense if you do 2b
And i don't think i did 4c properly =P
LostAuzzie
Member
For Question two you use the Cosine Rule and end up with, a^2, B^2 & a.b all in the one equation
AntiHyper
Revered Member
The markers for this assignment must be pretty lazy lol
i hope they picked the questions i'm doing well.
melimoo
knows how to rave
okay. how i love this thread
this is a last resort cause ive tried and really can't do any of the following from the math1001 assignment:
3b
4c
5
http://www.maths.usyd.edu.au/u/UG/JM/MATH1001/r/PDF/ass2.pdf
for 3b, i looked at how the limit behaves as you approach along x-axis, y-axis and the line x=y. and for all i got them to be = 0. essentially i was trying to show that the limit didn't exist, but to my understanding it doesn't exist if all those values aren't equal. so is the limit = 0? or does it exist? wah
4c, trying to sub in x = pi/2 and x= 3pi/2 and show that they are different signs. they both = 0
:/
5, no fucking idea
please help meeeeee thankyou <3
this is a last resort cause ive tried and really can't do any of the following from the math1001 assignment:
3b
4c
5
http://www.maths.usyd.edu.au/u/UG/JM/MATH1001/r/PDF/ass2.pdf
for 3b, i looked at how the limit behaves as you approach along x-axis, y-axis and the line x=y. and for all i got them to be = 0. essentially i was trying to show that the limit didn't exist, but to my understanding it doesn't exist if all those values aren't equal. so is the limit = 0? or does it exist? wah
4c, trying to sub in x = pi/2 and x= 3pi/2 and show that they are different signs. they both = 0
:/
5, no fucking idea
please help meeeeee thankyou <3
SeDaTeD
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For 3 b, let it approach the origin along y=kx, for some constant k.
for 4c, use part a, which says the limit as x-> pi/2 is 1, and that f(x) is essentially the same as the equation except for one point and that it is continuous. Sub in 3pi/2 then use intermediate value theorem.
q5, by MVT, you have, for some c, a<c<b, -sin(c) = [cos(b)-cos(a)]/[b-a]. But |-sin(c)| =< 1 for any c. so |[cos(b)-cos(a)]/[b-a]| =< 1. Rearrange to get the result.
for 4c, use part a, which says the limit as x-> pi/2 is 1, and that f(x) is essentially the same as the equation except for one point and that it is continuous. Sub in 3pi/2 then use intermediate value theorem.
q5, by MVT, you have, for some c, a<c<b, -sin(c) = [cos(b)-cos(a)]/[b-a]. But |-sin(c)| =< 1 for any c. so |[cos(b)-cos(a)]/[b-a]| =< 1. Rearrange to get the result.
melimoo
knows how to rave
thankyou. it helped...a little