Length and Angle of Vectors

[Petrus]

Hello MHB,
This is an old exam.
In top of the exam paper it says. Evrything is on Orthogonal-System

Vector \(\displaystyle u\) got the length \(\displaystyle 3\), vector \(\displaystyle v\) got the length \(\displaystyle 2\) and vector u-v got the length \(\displaystyle \sqrt{7}\)
(a) calculate length of vector \(\displaystyle u+v\)
(b) calculate the angle between \(\displaystyle u\) and \(\displaystyle v\)

I did not understand how you do (a) and for (b) I got confused cause if it's orthogonal System then the angle will be \(\displaystyle \frac{\pi}{2}\) but that is wrong. So I assume we don't got orthogonal system in this one?

Regards,
\(\displaystyle |\pi\rangle\)

 

[Poirot]

Re: Vector

Hello MHB,
This is an old exam.
In top of the exam paper it says. Evrything is on Orthogonal-System

Vector \(\displaystyle u\) got the length \(\displaystyle 3\), vector \(\displaystyle v\) got the length \(\displaystyle 2\) and vector u-v got the length \(\displaystyle \sqrt{7}\)
(a) calculate length of vector \(\displaystyle u+v\)
(b) calculate the angle between \(\displaystyle u\) and \(\displaystyle v\)

I did not understand how you do (a) and for (b) I got confused cause if it's orthogonal System then the angle will be \(\displaystyle \frac{\pi}{2}\) but that is wrong. So I assume we don't got orthogonal system in this one?

Regards,
\(\displaystyle |\pi\rangle\)

|u+v|^2=u.u+v.v+2u.v=9+4=13 so |u+v| is the square root of 13.

b) u.v=|u||v|cosx. Solve for x.

 

[Petrus]

Re: Vector

|u+v|^2=u.u+v.v+2u.v=9+4=13 so |u+v| is the square root of 13.

a) u.v=|u||v|cosx. Solve for x.

the answer says \(\displaystyle \sqrt{19}\) and in a ON system \(\displaystyle u*v=0\) so I assume they don't mean ON system, thanks for confirming that i don't misunderstand! Now I get same answer as the solution ( I wounder what hapend those who did write that exam and did 'misunderstand')
\(\displaystyle |u-v|^2=u*u-2uv+v*v \) and we know \(\displaystyle |u-v|=\sqrt{7}\) so \(\displaystyle u*v=3\)
Thanks allot, got almost crazy and start read over about ON system over and over again

(b) If we assume it was ON system it would be \(\displaystyle \frac{\pi}{2}\) that would be correct right? Or do I have misunderstand with this scalar product?

Regards,
\(\displaystyle |\pi\rangle\)

 

[Klaas van Aarsen]

Re: Vector

In an orthonormal system, you have a set of base vectors that are orthogonal and have length 1.
A vector in this system is a linear combination of the base vectors.
Two such vectors do not have to be orthogonal nor do they need to have length 1.

 

[Poirot]

Re: Vector

How can there be any ambiguity? An orthogonal set means the inner product of all distinct vectors is zero. The set is orthonormal if in addition, the magnitude of all vectors is one.