MHB S6.12.25 find v in component form

[karush]

$\tiny{s6.12.25}$
$\textsf{If $v$ lies in the first quarter and makes an angle }\\$
$\textsf{$\pi/3$ with the positive x-axis and $\left| v \right|$=4} $
$\textsf{find $v$ in component form.}$
\begin{align}
\displaystyle
v&=\langle 2\sqrt{3},2\rangle \\
\end{align}
this is probably correct, but always suggestions... notice that $\langle \rangle$ are not on the lateX menu

 

[topsquark]

$\tiny{s6.12.25}$
$\textsf{If $v$ lies in the first quarter and makes an angle }\\$
$\textsf{$\pi/3$ with the positive x-axis and $\left| v \right|$=4} $
$\textsf{find $v$ in component form.}$
\begin{align}
\displaystyle
v&=\langle 2\sqrt{3},2\rangle \\
\end{align}
$\textit{this is probably correct, but always suggestions... notice that $\langle \rangle$ are not on the lateX menu and word wrap does not function in Live preview!}$

You got your trig functions backward.

In component form we have that any vector v based at the origin has the form [math]( |v| \cdot cos( \theta ), |v| \cdot sin( \theta ) )[/math]. In your case:
[math]\left ( 4 \cdot cos \left ( \frac{\pi}{3} \right ), 4 \cdot sin \left ( \frac{\pi}{3} \right ) \right )[/math]

[math]= \left ( 4 \cdot \frac{1}{2} , 4 \cdot \frac{\sqrt{3}}{2} \right )[/math]

[math]= ( 2, 2 \sqrt{3} )[/math]

-Dan

Addendum: Your problem would seem to be in your "textit" line. If you just type it out the wrap works just fine.

 

[karush]

$\textsf{If $v$ lies in the first quarter and makes an angle $\pi/3$ with the positive x-axis and $\left| v \right|$=4 find $v$ in component form.}$
\begin{align}
\displaystyle
(a_1,b_1)&=\langle 4 \cdot cos \left ( \frac{\pi}{3} \right ), 4 \cdot sin \left ( \frac{\pi}{3} \right ) \rangle\\
v&=\langle 2\sqrt{3},2\rangle
\end{align}