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math121b:ex3 [2020/02/20 14:36]
pzhou
math121b:ex3 [2020/02/20 15:02] (current)
pzhou
Line 62: Line 62:
  (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/ex3.1582238182.txt.gz · Last modified: 2020/02/20 14:36 by pzhou