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[SOLVED] 311.1.3.2 u+v and u-2v and tikx?

karush

Well-known member
Jan 31, 2012
3,068
$\tiny{311.1.3.2}$
Compute $u+v$ and $u-2v$
$u=\left[
\begin{array}{rr} 3\\2 \end{array}\right], v=\left[
\begin{array}{rr}2\\-1 \end{array}\right]
\quad u+v=\left[\begin{array}{rr}3+2\\2-1 \end{array}\right]=
\left[\begin{array}{rr}5\\4 \end{array}\right]
\quad u-2v=\left[\begin{array}{rr}3-2(2)\\2-2(-1) \end{array}\right]
=\left[\begin{array}{rr}1\\0 \end{array}\right]$

ok I think this is correct typos maybe, but the next question is

Display the vectors using arrows on an xy-graph
$u,v, -v, -2v, u+v, u-v, $ and $u-2v$

I was going to try this with tikx but was looking for an example to follow since we use arrows

also is these vectors $\mathbb{R}^2$

Mahalo
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
1,007
$\tiny{311.1.3.2}$
Compute $u+v$ and $u-2v$
$u=\left[
\begin{array}{rr} 3\\2 \end{array}\right], v=\left[
\begin{array}{rr}2\\-1 \end{array}\right]
\quad u+v=\left[\begin{array}{rr}3+2\\2-1 \end{array}\right]=
\left[\begin{array}{rr}5\\4 \end{array}\right]
\quad u-2v=\left[\begin{array}{rr}3-2(2)\\2-2(-1) \end{array}\right]
=\left[\begin{array}{rr}1\\0 \end{array}\right]$

ok I think this is correct typos maybe, but the next question is

Display the vectors using arrows on an xy-graph
$u,v, -v, -2v, u+v, u-v, $ and $u-2v$

I was going to try this with tikx but was looking for an example to follow since we use arrows

also is these vectors $\mathbb{R}^2$

Mahalo
uhh ...

$u+v = \begin{bmatrix}
5\\ 1

\end{bmatrix}$

$u-2v = \begin{bmatrix}
-1\\ 4

\end{bmatrix}$

try again ...
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
I am hoping that you know, perfectly well, that 2- 1= 1, not 4, and, although it is slightly more complicated, that 2- 2(-1)= 4. not 0.
 

karush

Well-known member
Jan 31, 2012
3,068
$u=\left[
\begin{array}{rr} 3\\2 \end{array}\right], v=\left[
\begin{array}{rr}2\\-1 \end{array}\right]
\quad u+v=\left[\begin{array}{rr}3+2\\2-1 \end{array}\right]=
\left[\begin{array}{rr}5\\1 \end{array}\right]
\quad u-2v=\left[\begin{array}{rr}3-2(2)\\2-2(-1) \end{array}\right]
=\left[\begin{array}{rr}-1\\4 \end{array}\right]$
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
1,007
$u=\left[
\begin{array}{rr} 3\\2 \end{array}\right], v=\left[
\begin{array}{rr}2\\-1 \end{array}\right]
\quad u+v=\left[\begin{array}{rr}3+2\\2-1 \end{array}\right]=
\left[\begin{array}{rr}5\\1 \end{array}\right]
\quad u-2v=\left[\begin{array}{rr}3-2(2)\\2-2(-1) \end{array}\right]
=\left[\begin{array}{rr}-1\\4 \end{array}\right]$
So ... what now?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,274

karush

Well-known member
Jan 31, 2012
3,068
how do you write a vector in tikz
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
1,007
haven’t taken the time to learn tikz, but geogebra does a nice job and is user friendly ...

a = u+v
b = u-v
w = -v
d = -2v
c = u-2v


4976082C-B294-42FB-8901-39F84602264F.png
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
how do you write a vector in tikz
The following latex code does the job:
LaTeX:
\begin{tikzpicture}
  \draw[->] (0,0) -- (3,2);
\end{tikzpicture}
\begin{tikzpicture}
\draw[->] (0,0) -- (3,2);
\end{tikzpicture}

A prettier version with some embellishments is:
LaTeX:
\begin{tikzpicture}
  \coordinate[label=right:$\mathbf u$] (u) at (3,2);
  \draw[-latex, thick] (0,0) -- node[above left] {$\vec u$} (u);
\end{tikzpicture}
\begin{tikzpicture}
\coordinate[label=right:$\mathbf u$] (u) at (3,2);
\draw[-latex, thick] (0,0) -- node[above left] {$\vec u$} (u);
\end{tikzpicture}
 

karush

Well-known member
Jan 31, 2012
3,068
mahalo much
I'm trying to audit linear algrebra from UH west this spring but it will be via Google classroom so trying to get some early input

btw how do you get tikz to render on MHB

the reason I am using tikz is that will render in Overleaf which is commonly used here at UHW

Screenshot 2020-12-11 at 10.53.00 AM.png
 
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