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math121b:ex3 [2020/02/10 13:24]
pzhou
math121b:ex3 [2020/02/20 14:36]
pzhou
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 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 $\vec v_1, \vec v_2$. $$   P(\vec v_1, \vec v_2) = ? $$+  * 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** area ((Signed area means $P(v,w) = - P(w,v)$)) of the parallelogram (skewed rectangle) generated by $\vec v_1, \vec v_2$. $$   P(\vec v_1, \vec v_2) = ? $$
   * From the above computation, can you deduce $$ P(\vec v_1 + 3 \vec v_2, \vec v_2) = ?$$ which formula did you use?    * From the above computation, can you deduce $$ P(\vec v_1 + 3 \vec v_2, \vec v_2) = ?$$ which formula did you use? 
   * How about $$ P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=? $$   * How about $$ P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=? $$
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 ===== 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. 
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   * $ \| 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 $ 
-  * $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{4} = 2$?+  * $P(e_1, e_2) = \sqrt{\det(g)} = \sqrt{4} = 2$
   * 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 
 $$ v \cdot w = \| v \| \|w \| \cos \theta$$  $$ v \cdot w = \| v \| \|w \| \cos \theta$$ 
math121b/ex3.txt · Last modified: 2020/02/20 15:02 by pzhou