Prove that the angle between position vectors are equal

[consuela]

Hi,

I have three position vectors A=(3,-1,2) B=(-1,-1,-2) C=(3,3,-2) the question says to prove that these three vectors have the same angle I solved it by this law "Cos0 = a.b / |a||b|" and I found the angle between A and B , A and C, B and C as following

for A and B:
A.B = 3(-1)+-1(-1)+2(-2) = -6
|A| = √3² + -1² + 2² = √14

Same for |B|

cos0 = A.B / |A||B| = -6 /√14√6 = -√21 /7 -----> 0 = cos-1(-√21/7)= 130,89°

for A and C = as I did above same steps = 83.456°

for B and C = 150,50°

so my answers aren't match I have asked my teacher he told me to solve it by another way and I couldn't so I am here asking for help. Also I know I have to write it in radian that's not a problem I will convert it at the end

 

[lev888]

Hi,

I have three position vectors A=(3,-1,2) B=(-1,-1,-2) C=(3,3,-2) the question says to prove that these three vectors have the same angle I solved it by this law "Cos0 = a.b / |a||b|" and I found the angle between A and B , A and C, B and C as following

for A and B:
A.B = 3(-1)+-1(-1)+2(-2) = -6
|A| = √3² + -1² + 2² = √14

Same for |B|

cos0 = A.B / |A||B| = -6 /√14√6 = -√21 /7 -----> 0 = cos-1(-√21/7)= 130,89°

for A and C = as I did above same steps = 83.456°

for B and C = 150,50°

so my answers aren't match I have asked my teacher he told me to solve it by another way and I couldn't so I am here asking for help. Also I know I have to write it in radian that's not a problem I will convert it at the end

Wolfram Alpha confirms that angles are not the same. Are you sure you are calculating the right angles? I don't know what "prove that these three vectors have the same angle" even means. Vectors don't have angles.

 

[consuela]

Wolfram Alpha confirms that angles are not the same. Are you sure you are calculating the right angles? I don't know what "prove that these three vectors have the same angle" even means. Vectors don't have angles.

Well it's not the angle of a vector it self but the angle between two vectors as I found such as the angle between A and B and so on the teacher says it's the same angle between each but I can't prove that

 

[consuela]

Wolfram Alpha confirms that angles are not the same. Are you sure you are calculating the right angles? I don't know what "prove that these three vectors have the same angle" even means. Vectors don't have angles.

And yes I am 100% sure this answers are correct and I solved it by the law I write above I believe there is another way. I also tried the law contains sin0 and the answers not same too

 

[lev888]

And yes I am 100% sure this answers are correct and I solved it by the law I write above I believe there is another way. I also tried the law contains sin0 and the answers not same too

Then show your work to the teacher and ask for an explanation.

 

[consuela]

Then show your work to the teacher and ask for an explanation.

Well he told me to think over, and he is not that helpful teacher also this is part of homework if I didn't get it wright I may get F ... long topic can't you help me?

 

[lev888]

Well he told me to think over, and he is not that helpful teacher also this is part of homework if I didn't get it wright I may get F ... long topic can't you help me?

I don't see any mistakes in your work if we are indeed looking for the 3 angles between vectors.

 

[consuela]

Okay thank you for helping I will see if I could find something in the Internet

 

[Dr.Peterson]

Okay thank you for helping I will see if I could find something in the Internet

What more is there to look for? The problem as you state it is simply wrong.

To my mind, the thing to do is to make sure you are interpreting the problem correctly. Please show us the original of the problem so we can make sure we've got the right one.

 

[consuela]

Hi, I have found the ans it's a little tricky you can see be calculating the disance between each vectors A and B ,, A and C ,, B and C you will find it's the same distance and by connecting them they will form a triangular with same sides -I don't know what is called in English anyway- and have the same angles tddaaaa thaks to some one in mathoverflow helped me and thank you guys too I really appreciate your time

 

[lev888]

Hi, I have found the ans it's a little tricky you can see be calculating the disance between each vectors A and B ,, A and C ,, B and C you will find it's the same distance and by connecting them they will form a triangular with same sides -I don't know what is called in English anyway- and have the same angles tddaaaa thaks to some one in mathoverflow helped me and thank you guys too I really appreciate your time

Then A, B, C are not vectors, they are points. You are asked to prove that angles of the triangle ABC are equal. Posting the exact text of the problem (or at least google translate) would've helped.

 

[pka]

Hi, I have found the ans it's a little tricky you can see be calculating the disance between each vectors A and B ,, A and C ,, B and C you will find it's the same distance and by connecting them they will form a triangular with same sides -I don't know what is called in English anyway- and have the same angles tddaaaa thaks to some one in mathoverflow helped me and thank you guys too I really appreciate your time

As Lev888 said \(A,~B~\&~C\) are points and \(\overrightarrow {AB} = \left\langle { - 4,0, - 4} \right\rangle ,\quad \overrightarrow {BC} = \left\langle {4,4,0} \right\rangle ,{\kern 1pt} {\kern 1pt} \overrightarrow {CA} = \left\langle {0, - 4,4} \right\rangle \)
Each of those vectors has length \(\sqrt{32}\) hence the \(\Delta ABC\) is equilateral and each angle measures \(\dfrac{\pi}{3}\)

 

[Dr.Peterson]

I have three position vectors A=(3,-1,2) B=(-1,-1,-2) C=(3,3,-2) the question says to prove that these three vectors have the same angle I solved it by this law "Cos0 = a.b / |a||b|" and I found the angle between A and B , A and C, B and C as following

Do you see how your wording was incorrect? You describe the points as "position vectors", and then refer to "these three vectors", which implies you mean the position vectors, not the vectors AB, BC, and CA (or whatever). I really want to see the original problem!

"Have the same angle" is also bad wording, as has already been pointed out.

 

[pka]

"Have the same angle" is also bad wording, as has already been pointed out.

That is a truly generous understatement.

 

[consuela]

Hi, yes I went back to teacher and told me that this answer too is wrong the question was written good there is no problem with it I believe there is a mistake with the question as you say

 

[pka]

Hi, yes I went back to teacher and told me that this answer too is wrong the question was written good there is no problem with it I believe there is a mistake with the question as you say

Why don't you post the question in the original language?

 

[Dr.Peterson]

Hi, yes I went back to teacher and told me that this answer too is wrong the question was written good there is no problem with it I believe there is a mistake with the question as you say

Let's suppose that the problem really meant this:

Given the position vectors A=(3,-1,2) B=(-1,-1,-2) C=(3,3,-2), prove that the angles between vectors AB and AC, between vectors BA and BC, and between angles CA and CB are equal (that is, the angles in the triangle ABC are equal).​

Then, as pka said in post #12, it would be making a true statement that you can prove.

Do you see that vector AB = <-4, 0, -4>, BC = <4, 4, 0>, and CA = <0, -4, 4>?

Then, for example, the angle between AB and AC is the angle between <-4, 0, -4> and <0, 4, -4>; AB.AC = 0+0+16 = 16; |AB| = 4sqrt(2); |AC| = 4sqrt(2); so cos(theta) = 16/(4sqrt(2)*4sqrt(2)) = 16/32 = 1/2, so theta = 60 degrees; and so on.

So the only thing missing is to see, as we have asked several times, how the question is actually worded (in the original language), so we can help you see why it means this.