This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
math121b:ex3 [2020/02/10 13:24] pzhou |
math121b:ex3 [2020/02/20 15:02] (current) pzhou |
||
---|---|---|---|
Line 12: | Line 12: | ||
2. Area | 2. Area | ||
- | * Consider the vector space $\R^2$. Let $\vec v_1 = (1,1), \vec v_2 = (0,2)$. Let $P(\vec v_1, \vec v_2)$ denote the area of the parallelogram (skewed rectangle) generated by $\vec v_1, \vec v_2$. $$ | + | * Consider the vector space $\R^2$. Let $\vec v_1 = (1,1), \vec v_2 = (0,2)$. Let $P(\vec v_1, \vec v_2)$ denote the **signed** |
* From the above computation, | * From the above computation, | ||
* How about $$ P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=? $$ | * How about $$ P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=? $$ | ||
Line 43: | Line 43: | ||
===== Solution ===== | ===== Solution ===== | ||
+ | ==== Metric Tensor. Length, Area and Volume element ==== | ||
1. True of False | 1. True of False | ||
* False. Although, you can equip any finite dimensional vector space with an inner product, there is no one holier than the other. | * False. Although, you can equip any finite dimensional vector space with an inner product, there is no one holier than the other. | ||
Line 60: | Line 61: | ||
* $ \| e_1 + e_2 \|^2 = g(e_1, e_1) + g(e_2, e_2) + 2 g(e_1, e_2) = 3 + 2 + 2 = 7 $, hence $\| e_1 + e_2 \| = \sqrt{7} $. Another way to get $\| e_1 + e_2 \|^2$ is by | * $ \| e_1 + e_2 \|^2 = g(e_1, e_1) + g(e_2, e_2) + 2 g(e_1, e_2) = 3 + 2 + 2 = 7 $, hence $\| e_1 + e_2 \| = \sqrt{7} $. Another way to get $\| e_1 + e_2 \|^2$ is by | ||
$$ \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7 $$ | $$ \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7 $$ | ||
- | * $ \| e_1 - 2 e_2 \| = \begin{pmatrix} 1 & -2 \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr -2 \end{pmatrix} = 7 $ | + | * $ \| e_1 - 2 e_2 \|^2 = \begin{pmatrix} 1 & -2 \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr -2 \end{pmatrix} = 7 $ |
- | * $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{4} = 2$? | + | * $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{5}$ |
* Can you find two vectors $v_1, v_2 \in \R^2$, such that $v_1, v_2$ has the same properties as $e_1, e_2$? $v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta)$ where $\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}$. We are using the formula | * Can you find two vectors $v_1, v_2 \in \R^2$, such that $v_1, v_2$ has the same properties as $e_1, e_2$? $v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta)$ where $\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}$. We are using the formula | ||
$$ v \cdot w = \| v \| \|w \| \cos \theta$$ | $$ v \cdot w = \| v \| \|w \| \cos \theta$$ |