Consider the rectangle formed by the points $(2,7),(2,6),(4,7)$ and
$(4,6)$. Is it still a rectangle after transformation by $\underline
A$= $ \left( \begin{matrix} 3&1 \\ 2&\frac {1}{2} \\ \end{matrix}
\right) $ ?By what factor has its area changed ?
I've defined the point $(2,6)$ as the origin of my vectors $\vec v $ and $\vec u$ with
$\vec v = \left(\begin{matrix}0 \\1 \\\end{matrix} \right)$ and $\vec u = \left(\begin{matrix}2 \\0 \\\end{matrix} \right)$ which get transformed to $\underline A \vec v=$$\left(\begin{matrix}1 \\\frac{1}{2} \\\end{matrix} \right)$ and $\underline A \vec u=$$\left(\begin{matrix}6 \\4 \\\end{matrix} \right)$.
So my new figure(which is not a rectangle anymore,but is now a parallelogram) has vertices $(2,6)(3,6 \frac{1}{2}),(8,10)$ and $(9,10 \frac{1}{2})$
Now the rectangle has area equal to $2 \cdot 1=2$, and after the transformation I have that the area of the resulting parallelogram is $\underline A \vec v \times \underline A \vec u =|1\cdot 4 -\cfrac{1}{2}\cdot 6|=1$
Now my problem is that when I calculate the area by geometric methods I have:
You see I get a different answer,so it's clear that I've had it all wrong since the beginning but I don't see where.
I upload now the image of the parallelogram where I've applied law of cosines in the last step of the above image.
I've tried to be as specific as possible about my steps.Can someone help me ?
Thanks in advance.
Best Answer
This is just the cosine rule:
$\cos \theta = {\|u\|^2 +\|v\|^2 - \|u-v\|^2 \over 2 \|u\| \|v\| } = {52 +{5 \over 4} - {149 \over 4} \over 2 \sqrt{ 52 } \sqrt{{5 \over 4} } }$.
You are missing a square root or two.