exercise:981da7a5df: Difference between revisions

From Stochiki
(Created page with "<div class="d-none"><math> \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow}...")
 
No edit summary
 
Line 31: Line 31:
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
\newcommand{\mathds}{\mathbb}
\newcommand{\dilemma}[2] {
\left\{
\begin{array}{ll}
#1 \\
#2
\end{array}
\right.
}
\newcommand{\trilemma}[3] {
\left\{
\begin{array}{ll}
#1 \\
#2 \\
#3
\end{array}
\right.
}
</math></div>
</math></div>
<ul style{{=}}"list-style-type:lower-alpha"><li></li>
<ul style{{=}}"list-style-type:lower-alpha">
<li>lab{10.5.1a}
<li>
Draw each of the following vectors.
Draw each of the following vectors.
(i) <math>(0,5)_{P_0}</math>, where <math>P_0 = (-1,1)</math>.
(i) <math>(0,5)_{P_0}</math>, where <math>P_0 = (-1,1)</math>.
Line 40: Line 57:
(iv) <math>(-2,-3)_{P_3}</math>, where <math>P_3 = (0,0)</math>.</li>
(iv) <math>(-2,-3)_{P_3}</math>, where <math>P_3 = (0,0)</math>.</li>
<li>Let <math>P_0 = (-1,1)</math>, and compute and draw the
<li>Let <math>P_0 = (-1,1)</math>, and compute and draw the
translated vectors <math>T_{P_0}(\vec u)</math>,
translated vectors <math>T_{P_0}(\vec u)</math>, where <math>\vec u</math> is taken to be each of the four vectors in (a).</li>
where <math>\vec u</math> is taken to be each of the four
vectors in \ref{ex10.5.1a}.</li>
</ul>
</ul>

Latest revision as of 22:52, 25 November 2024

[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} \newcommand{\dilemma}[2] { \left\{ \begin{array}{ll} #1 \\ #2 \end{array} \right. } \newcommand{\trilemma}[3] { \left\{ \begin{array}{ll} #1 \\ #2 \\ #3 \end{array} \right. } [/math]
  • Draw each of the following vectors. (i) [math](0,5)_{P_0}[/math], where [math]P_0 = (-1,1)[/math]. (ii) [math](4,-1)_{P_1}[/math], where [math]P_1 = (1,-1)[/math]. (iii) [math](1,3)_{P_2}[/math], where [math]P_2 = (1,1)[/math]. (iv) [math](-2,-3)_{P_3}[/math], where [math]P_3 = (0,0)[/math].
  • Let [math]P_0 = (-1,1)[/math], and compute and draw the translated vectors [math]T_{P_0}(\vec u)[/math], where [math]\vec u[/math] is taken to be each of the four vectors in (a).