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Mathematics 25 Online
OpenStudy (anonymous):

find the measure of the angle ABC. given: A(3,0,0) B(0,1,0) C(1,2,3)

OpenStudy (angela210793):

Perimeter?

OpenStudy (anonymous):

no have to find the angle :)

OpenStudy (angela210793):

hmm....let me see a formula..I've 4gotten it :(

OpenStudy (anonymous):

cosθ=(x1x2+y1y2+z1z2)/|u||v|

OpenStudy (anonymous):

thats what i used but it could be the wrong formula

OpenStudy (angela210793):

(a1a2a3+b1b2b3) the angle between 2 lines is cosA =------------------------------------ sqr(a1^2+b1^2+c1^2)*sqr(a2^2+b2^2+c2^2) k1k2+1 Or cosA=--------------------------- sqr(k1^2+1*sqr(k2^2+1)

OpenStudy (angela210793):

too messy

OpenStudy (anonymous):

yes i used the same formula i dont know why my answer came out wrong

OpenStudy (anonymous):

is the angle between vector AB and vector AC correct

OpenStudy (angela210793):

i thought u might have to find the eq of the lines and then use the coefficients or slopes...

OpenStudy (anonymous):

AB(-3,1,0) and AC(-2,2,3)

OpenStudy (anonymous):

i dont think so. But i could be wrong

OpenStudy (angela210793):

Or i can be wrong...

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

omg this is killing me:)

OpenStudy (angela210793):

I don't have the idea how to do it...:S:S ...Sorry..:(:( Survive till some1 else helps u though :P:P

OpenStudy (anonymous):

lol ill try

OpenStudy (angela210793):

^_^ lol

myininaya (myininaya):

this is what i got

myininaya (myininaya):

and have you opened it?

OpenStudy (anonymous):

yep just a quick question. when you find the vector AB shouldnt it be the vector b-a?

myininaya (myininaya):

you know i was never sure about that lol you are probably right i never had a class on this, but i have looked at it

myininaya (myininaya):

AB and BA have different direction so they are different vectors right?

OpenStudy (anonymous):

thats correct

myininaya (myininaya):

i know we want to find angle B so I know I'm suppose to use CB/BC and AB/BA but i'm not sure what to use maybe I think i;m suppose to use AB and BC

OpenStudy (anonymous):

yes i was thinking the same the thing is when i use AB and AC my answer is wrong lol

myininaya (myininaya):

AB=<0-3,1-0,0-0> BC=<1-0,2-1,3-0>

myininaya (myininaya):

AB dot BC=-3(1)+1(1)+0(3)=-3+1=-2 mag{AB}=sqrt{9+1}=sqrt{10} mag{BC}=sqrt{1+1+9}=sqrt{11}

myininaya (myininaya):

\[\cos^{-1}(\frac{-2}{\sqrt{110}})\]

myininaya (myininaya):

what do you think of that?

OpenStudy (anonymous):

the answer is \[\approx 79.01 degrees\]

OpenStudy (anonymous):

so your working out before was closer the answer

myininaya (myininaya):

omg the first one i had was right

OpenStudy (anonymous):

lol sorry

OpenStudy (anonymous):

so its BC and BA

myininaya (myininaya):

so yes i labeled them wrong that is so weird how i didn't bother with the directions and i wounded up doing the right way with wrong labels

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

thanks heaps :)

myininaya (myininaya):

thanks for teaching me alittle bit about vectors

OpenStudy (angela210793):

Congratz guys :)

myininaya (myininaya):

by the way and you called me a guy earlier i am not lol

myininaya (myininaya):

using BC and BA make sense since we listed the angle we looking for first that angle being B

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