Question

Let vector AB=(6,-2,3)and vector AC= (-2,-3,2). How do I find BÂC?

Answer 1

AB+BC+CA=0

Answer 2

do you understand why?

Answer 3

no

Answer 4

draw a triangle. Label the three vertices A B and C. See that if you move from A to B, then from B to C, then from C to A, you end up right where you started: at A.

Answer 5

.ok I see that.

Answer 6

so the total vector change (going from A to B, from B to C, and from C back to A) AB+BC+CA=0 (you end exactly where you started)

Answer 7

OH. OK I understand,but how do I substitute the coordinates into that equation?

Answer 8

you have AB and AC.
Look back at your triangle. If you move from A to C and then from C back to A, you end up right where you started, right? so AC + CA = 0, so AC = -CA

Answer 9

so then you have AB and CA. Substitute them into the equation and you can find BC by algebra.

Answer 10

hold on, I just realize I wrote the problem wrong. I'm trying to find angle "a" not vector BC. I'm sorry

Answer 11

do you know how to take the dot product of two vectors?

Answer 12

no, my teacher have not taught my class how to multiply vectors. we just started. *BÂC

Answer 13

the easy way to find the angle between the two sides is with the dot product. The more complicated way uses the Law of Cosines.

Answer 14

I know the Law of cosine. we just quizzed on that and law of sine.

Answer 15

so to find the angle with law of cosines, find BC, find the magnitudes of all three vectors, and set up the triangle. The side lengths are the magnitudes. You can then use the law of cosines to find the angle

Answer 16

How would you find the magnitudes?

Answer 17

it's kind of like using the pythagorean theorem twice. Once to find out the magnitude of just the x coordinate and the y coordinate, then again to find out the magnitude of the magnitude you found before and the z coordinate.
\[distance=\sqrt{\sqrt{x^2+y^2}^2+z^2}\]simplifies to
\[distance=\sqrt{x^2+y^2+z^2}\]

Answer 18

(sorry it took so long)

Answer 19

no I'm sorry I keeping looking at the wrong problem. I am trying to find vector BC. I'm so sorry I making you all this work. I'm such a mess.

Answer 20

it's fine

Answer 21

I would not be surprise if you stop helping me . I understand

Answer 22

you already know how to do this, I showed you up above.

Answer 23

AB+BC+CA=0
just use algebra

Answer 24

I meant to write how do I substitute AB and AC into the equation?

Answer 25

CA is -AC, so CA is (2,3,-2)

Answer 26

has your teacher taught you how to add vectors?

Answer 27

I understand that. When I substitution the coordinates, how would I write it. That is the part where I am confused at. She just began teacher us vectors, like a seven min intro. So, my class is basically got nowhere.

Answer 28

she like and always makes us figure it out ourselves.

Answer 29

I write vectors either as
\[\left(\begin{matrix}x \\ y \\ z \end{matrix}\right)\]
or as (x,y,z)

Answer 30

ok

Answer 31

the first form is better if you want other people to understand it, but the second form is faster and takes less paper

Answer 32

well, which way do u want to do it?

Answer 33

it's your math problem, you choose how you write it
but your teacher would probably be happier if you wrote it vertically
so you have \[\left(\begin{matrix}6 \\ -2 \\ 3\end{matrix}\right) + \left(\begin{matrix}2 \\ 3 \\ -2\end{matrix}\right)=-BC\]

Answer 34

ok

Answer 35

so, does BC equal (-8,-1,-1)?

Answer 36

looks like it does

Answer 37

yah. thankyou so much!