1. ## Vector confusion.

Ok,
When you are finding an angle between two vectors, does it matter what direction their going in? I didn't think it did did...i mean an angle between two lines is an angle.
So this was the question.
ABCD is a parallelogram. Has position vectors A, B , C, D.
A =(2, 1, 3) B = (3, -2, 0) c = (1, -1, -2)
Determine to the nearest angle, ABC.
So need to use formula a.b = /a//b/Cos C (/a/ means for modulus, not dividing)
Where AB = a and BC = b
So i worked out
AB as (1, -3, -3)
and
BC as (-2, 1, -2)
a.b = -2 -3 + 6 = 1
/a/ = root 19
/b/ = 3
$cos c = \frac{1}{3\sqrt{19}}$
and therefore c = 86 degrees.
Well it was wrong and the answer was 94 degrees...and in the answer book it said i needed to do vector CB to get
$cos c = \frac{-1}{3\sqrt{19}}$
My question is why? Why does it matter if i use BC or CB and more importantly how will i know in an exam which one to use? No diagram was given...=(

Thanks!

2. Originally Posted by AshleyT
Ok,
When you are finding an angle between two vectors, does it matter what direction their going in? I didn't think it did did...i mean an angle between two lines is an angle.
So this was the question.

So need to use formula a.b = /a//b/Cos C (/a/ means for modulus, not dividing)
Where AB = a and BC = b
So i worked out
AB as (1, -3, -3)
and
BC as (-2, 1, -2)
a.b = -2 -3 + 6 = 1
/a/ = root 19
/b/ = 3
$cos c = \frac{1}{3\sqrt{19}}$
and therefore c = 86 degrees.
Well it was wrong and the answer was 94 degrees...and in the answer book it said i needed to do vector CB to get
$cos c = \frac{-1}{3\sqrt{19}}$
My question is why? Why does it matter if i use BC or CB and more importantly how will i know in an exam which one to use? No diagram was given...=(

Thanks!
Draw a diagram (2-D) will do showing three points and the relevant vectors and their directions. Now the inner (dot) product of the unit vectors in those directions will give you the cosine of the angle between the vectors. Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have.

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3. Originally Posted by Constatine11
Draw a diagram (2-D) will do showing three points and the relevant vectors and their directions. Now the inner (dot) product of the unit vectors in those directions will give you the cosine of the angle between the vectors. Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have.

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how will i know the directions of the vectors? There a clue in the question?

4. Originally Posted by AshleyT
how will i know the directions of the vectors? There a clue in the question?
The two rays of interest are BA and BC as B is the common point

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5. Originally Posted by Constatine11
The two rays of interest are BA and BC as B is the common point

.
Yes, and that gives the angle AshleyT originally got. What reason do you have for saying "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have"? Why is that supplementary angle the angle ABC?

6. Originally Posted by HallsofIvy
Yes, and that gives the angle AshleyT originally got. What reason do you have for saying "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have"? Why is that supplementary angle the angle ABC?
1. The OP used the vectors AB and and BC, not BA and BC.

2. "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have" that "which you did have" refers to the angle in the previous sentence. The angle we would have had from the proevous sentence was ABC, hence the supplementary angle is 180-ABC, which is what the OP had.

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7. Originally Posted by Constatine11
1. The OP used the vectors AB and and BC, not BA and BC.

2. "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have" that "which you did have" refers to the angle in the previous sentence. The angle we would have had from the proevous sentence was ABC, hence the supplementary angle is 180-ABC, which is what the OP had.

XX XXX XXX XXXX XX

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Yea i know thats what you do ,
But what i mean is, (sorry to annoy) but why do change...why do you find the supplementary angle and not the acute angle? The question never said 'find supplementary angle' or 'acute angle'.
How do you know its the supplementary angle that your mean to find, how do you know that?

Thanks

8. Originally Posted by AshleyT
Yea i know thats what you do ,
But what i mean is, (sorry to annoy) but why do change...why do you find the supplementary angle and not the acute angle? The question never said 'find supplementary angle' or 'acute angle'.
How do you know its the supplementary angle that your mean to find, how do you know that?

Thanks
Did you draw a diagram?

One angle is that between vectors AB and BC and the other is the angle between vectors BA and BC.

(The angle in question is the angle between the unit vectors when then are drawn pointing away from a common point)

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9. Originally Posted by Constatine11
1. The OP used the vectors AB and and BC, not BA and BC.
Oh, my goodness, I missed that completely! Thanks.

2. "Change the sign of one of the vactors and you are dealing with the suplementary angle to that which you did have" that "which you did have" refers to the angle in the previous sentence. The angle we would have had from the proevous sentence was ABC, hence the supplementary angle is 180-ABC, which is what the OP had.

XX XXX XXX XXXX XX

.

10. Originally Posted by Constatine11
Did you draw a diagram?

One angle is that between vectors AB and BC and the other is the angle between vectors BA and BC.

(The angle in question is the angle between the unit vectors when then are drawn pointing away from a common point)

.
Thank-you very, very much!! For all patience and replies