2
$\begingroup$

The question:

Two vectors $S=(\cos A,\cos B,\cos C)$, $S'=(\cos A',\cos B',\cos C')$, What is the angle between them?

The answer is $\cos(\theta)$ = $\cos A.\cos A'+ \cos B.\cos B'+ \cos C.\cos C'$.

Where does this answer come from??

It seems the two vectors are not unit vectors.

$\endgroup$
1

2 Answers 2

Reset to default
1
$\begingroup$

If you are given no restrictions on the values of $A, B, C, A', B', C'$, then I would agree that the answer they give is incorrect. It needs to be divided by the length of the vectors $(\cos A, \cos B, \cos C)$ and $(\cos A', \cos B', \cos C')$. As you say, if they are not unit vectors then there is a problem with their answer.

Perhaps there is some further information somewhere about the possible values of $A, B, C, A', B', C'$?

$\endgroup$
2
  • $\begingroup$ Thank you. There's no further infomation so I am pretty confused. It's actually from Theoretical Physics By Georg Joos. Exercise 5. $\endgroup$
    – Allitee
    Dec 6, 2013 at 12:26
  • $\begingroup$ I am sure the book is wrong. Maybe he missed out the words "unit vectors" in the question, or made a typo in the answers. If we take $A, B, \dots$ all to be zero, then the lengths of the two vectors is $\sqrt{3}$ and the given answer is definitely wrong. $\endgroup$
    – Old John
    Dec 6, 2013 at 12:35
0
$\begingroup$

Hint: $cos(\theta)= \frac{\langle{a,b}\rangle}{|a||b|}$

What do you get?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .