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--- math121b:ex3 [2020/02/10 13:24]
pzhou
+++ math121b:ex3 [2020/02/20 15:02] (current)
pzhou
@@ Line 12: / Line 12: @@
 . 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?
   * How about  $P(a \vec v_1 + b \vec v_2, c \vec v_1 + d \vec v_2)=?$
@@ Line 61: / 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
  $\begin{pmatrix} 1 & 1  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr 1 \end{pmatrix} = 7$
-  *  $\| e_1 - 2 e_2 \| =  \begin{pmatrix} 1 & -2  \end{pmatrix} \begin{pmatrix} 3 & 1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \cr -2 \end{pmatrix} = 7$
+  * $ \| 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
  $v \cdot w = \| v \| \|w \| \cos \theta$

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