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math121b:ex3 [2020/02/11 17:31]
pzhou
math121b:ex3 [2020/02/20 15:02] (current)
pzhou
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 2. Area 2. Area
-  * Consider the vector space R2\R^2. Let v1=(1,1),v2=(0,2)\vec v_1 = (1,1), \vec v_2 = (0,2). Let P(v1,v2)P(\vec v_1, \vec v_2) denote the area of the parallelogram (skewed rectangle) generated by v1,v2\vec v_1, \vec v_2.    P(\vec v_1, \vec v_2) = ? +  * Consider the vector space R2\R^2. Let v1=(1,1),v2=(0,2)\vec v_1 = (1,1), \vec v_2 = (0,2). Let P(v1,v2)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 v1,v2\vec v_1, \vec v_2.    P(\vec v_1, \vec v_2) = ?
   * From the above computation, can you deduce P(v1+3v2,v2)=? P(\vec v_1 + 3 \vec v_2, \vec v_2) = ? which formula did you use?    * From the above computation, can you deduce P(v1+3v2,v2)=? P(\vec v_1 + 3 \vec v_2, \vec v_2) = ? which formula did you use? 
   * How about P(av1+bv2,cv1+dv2)=? P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=?    * How about P(av1+bv2,cv1+dv2)=? P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=?
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  (11 )(3112)(11)=7  \begin{pmatrix} 1 & 1  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7   (11 )(3112)(11)=7  \begin{pmatrix} 1 & 1  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7
   * e12e22= (12 )(3112)(12)=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    * e12e22= (12 )(3112)(12)=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 v1,v2R2v_1, v_2 \in \R^2, such that v1,v2v_1, v_2 has the same properties as e1,e2e_1, e_2? v1=(3,0),v2=(2cosθ,2sinθ)v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta) where cosθ=123\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}. We are using the formula    * Can you find two vectors v1,v2R2v_1, v_2 \in \R^2, such that v1,v2v_1, v_2 has the same properties as e1,e2e_1, e_2? v1=(3,0),v2=(2cosθ,2sinθ)v_1 = (\sqrt{3}, 0), v_2 = (\sqrt{2} \cos\theta, \sqrt{2} \sin \theta) where cosθ=123\cos\theta = \frac{1}{\sqrt{2} \sqrt{3}}. We are using the formula 
 vw=vwcosθ v \cdot w = \| v \| \|w \| \cos \theta  vw=vwcosθ v \cdot w = \| v \| \|w \| \cos \theta 
math121b/ex3.1581471113.txt.gz · Last modified: 2020/02/11 17:31 by pzhou