Question

Please write the answer to all 4 uploading problems including this one on the paper and upload the answer in one pdf file after you finish and submit this final. [13].: - [] (9 points) Letü Recall that each vector correspond to a point in R2 (a). (3 points) Show that the triangle with vertices ő, ü, ö is a equilateral triangle (ie. a triangle with equal-length sides). What is the length of the three sides? (Hint: the third side can be expressed by vector ü - 7.) (b). (3 points) Show that {ü + jöis the height of this triangle w..t. the third side. i.e. show that yü + Lüis orthogonal to ü - . (c). (3 points) Can you find a pair of vectors ā, 6 s.t. the triangle with vertex o, a, b is a equilateral triangle with side-length 1?

Answer 1

$a)\overrightarrow{u}=\begin{bmatrix} 1\\ \sqrt{3} \end{bmatrix}\;\;,\;\;\overrightarrow{v}=\begin{bmatrix} 2\\ 0 \end{bmatrix}$

$\overrightarrow{0}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

We should prove that triangle with vertices form a equilateral triangle.

0,7,7

$\overrightarrow{0}-\overrightarrow{u}=\begin{bmatrix} 0\\ 0 \end{bmatrix}-\begin{bmatrix} 1\\ \sqrt{3} \end{bmatrix}=\begin{bmatrix} -1\\ -\sqrt{3} \end{bmatrix}$

$|\overrightarrow{0}-\overrightarrow{u}|=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2$

7-7= 1 13 2 0 -1 13

$|\overrightarrow{u}-\overrightarrow{v}|=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}}=2$

$\overrightarrow{v}-\overrightarrow{0}=\begin{bmatrix} 2\\ 0 \end{bmatrix}-\begin{bmatrix} 0\\ 0 \end{bmatrix}=\begin{bmatrix} 2\\ 0 \end{bmatrix}$

ū -= (22) +02 = V1 = 2

2-1 = 12-2 E-21 u

$\therefore\;triangle\;formed\;with\;vertices\;\overrightarrow{0},\overrightarrow{u},\overrightarrow{v}\;is\;an\;equilateral\;triangle.$

$b)\frac{1}{2}\overrightarrow{u}=\begin{bmatrix} \frac{1}{2}\\\frac{\sqrt{3}}{2} \end{bmatrix}\;\;,\;\;\frac{1}{2}\overrightarrow{v}=\begin{bmatrix} \frac{2}{2}\\ \frac{0}{2} \end{bmatrix}$

$\frac{1}{2}\overrightarrow{u}+\frac{1}{2}\overrightarrow{v}=\begin{bmatrix} \frac{1}{2}+1\\ \frac{\sqrt{3}}{2}+0 \end{bmatrix}$

$\frac{1}{2}\overrightarrow{u}+\frac{1}{2}\overrightarrow{v}=\begin{bmatrix} \frac{3}{2}\\ \frac{\sqrt{3}}{2} \end{bmatrix}$

$let\;\frac{1}{2}\overrightarrow{u}+\frac{1}{2}\overrightarrow{v}=\frac{3}{2}\hat{x}+\frac{\sqrt{3}}{2}\hat{y}$

$\overrightarrow{u}-\overrightarrow{v}=\begin{bmatrix} 1-2\\ \sqrt{3} -0\end{bmatrix}=\begin{bmatrix} -1\\ \sqrt{3} \end{bmatrix}$

$Let\;\overrightarrow{u}-\overrightarrow{v}=-1\hat{x}+\sqrt{3}\hat{y}$

$If\;(\frac{1}{2}\overrightarrow{u}+\frac{1}{2}\overrightarrow{v})*(\overrightarrow{u}-\overrightarrow{v})=0\;then\;the\;vectors\;are\;orthogonal.$

$(\frac{1}{2}\overrightarrow{u}+\frac{1}{2}\overrightarrow{v})*(\overrightarrow{u}-\overrightarrow{v})=(\frac{3}{2}\hat{x}+\frac{\sqrt{3}}{2}\hat{y})*(-1\hat{x}+\sqrt{3}\hat{y})$

$=\frac{3}{2}*(-1)+\frac{\sqrt{3}}{2}\sqrt{3}$

  $=-\frac{3}{2}+\frac{3}{2}$

$\therefore(\frac{1}{2}\overrightarrow{u}+\frac{1}{2}\overrightarrow{v})*(\overrightarrow{u}-\overrightarrow{v})are\;orthogonal.$

$c)Let\;th\;vertex\;\overrightarrow{0}=0\hat{i}+0\hat{j}$

$\overrightarrow{0},\overrightarrow{a},\overrightarrow{b}\;form\;a\;equilateral\;triangle\;with\;side\;length\;1.$

$\Rightarrow |\overrightarrow{0}-\overrightarrow{a}|=1,|\overrightarrow{a}-\overrightarrow{b}|=1,|\overrightarrow{b}-\overrightarrow{0}|=1$

$\overrightarrow{a}=a\hat{x}+b\hat{y}\;,\;\overrightarrow{b}=c\hat{x}+d\hat{y}$

$|\overrightarrow{0}-\overrightarrow{a}|=1\Rightarrow \sqrt{(0-a)^2+(0-b)^2}=1\Rightarrow \sqrt{a^2+b^2}=1\Rightarrow a^2+b^2=1$

$\overrightarrow{a}-\overrightarrow{b}=1\Rightarrow \sqrt{(a-c)^2+(b-d)^2}=1\Rightarrow \sqrt{a^2+c^2-2ac+b^2+d^2-2bd}=1$

$\Rightarrow {a^2+c^2-2ac+b^2+d^2-2bd}=1$

$\overrightarrow{b}-\overrightarrow{0}=1\Rightarrow \sqrt{(c-0)^2+(d-0)^2}=1\Rightarrow \sqrt{c^2+d^2}=1\Rightarrow c ^2+d^2=1$