Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I wanted to know if I did this problem right or not.

$A$ and $B$ are the following matrices:

$A=\begin{bmatrix}2&6\\1&-3\end{bmatrix},$

$B=\begin{bmatrix}3&2\\1&0\end{bmatrix}.$

Then

$$\frac{\langle A,B\rangle}{\|A\|\|B\|} =\frac{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix}, \begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}{\sqrt{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix},\begin{bmatrix}2&6\\1&-3\end{bmatrix}\right\rangle}\sqrt{\left\langle\begin{bmatrix}3&2\\1&0\end{bmatrix},\begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}}.$$

I ended up getting:

$\arccos(19/184) = 84.07^\circ$

share|improve this question
    
Justin: When I made the math prettier, I also took the liberty of correcting your denominator. I hope this wasn't going too far. I figured that it was a typo, because the expression you had was equal to $1$. But perhaps that was part of the problem. –  Jonas Meyer Apr 5 '11 at 1:37
add comment

1 Answer

up vote 0 down vote accepted

$a\cdot b=|a||b|\cos\theta$ so $$\theta=\arccos\Big(\frac{a\cdot b}{|a||b|}\Big).$$ we have $a\cdot b=6+12+1=19, |a||b|=\sqrt{4+36+1+9}\sqrt{9+4+1}=10\sqrt{7}$.
so the angle is $\arccos(19/(10\sqrt{7}))$

share|improve this answer
    
It's amazing that you could understand Justin's question. –  GWu Apr 5 '11 at 1:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.