hello!!
i'm having a little trouble converting between vector forms!!
so how would you convert the vector V = (3,40 degrees) into cartesian form
and convert v = 2i + 5j into polar form...
thanks!!
Hello, sir nerdalot!
How would you convert the vector $\displaystyle v \:= \:(3,\,40^o)$ into cartesian form
and convert $\displaystyle v \:= \:2i + 5j$ into polar form?
If you sketch this diagram, the pieces should come together.We can see that: .$\displaystyle \begin{array}{c}x \: =\: r\cos\theta \\ y \: =\: r\sin\theta\end{array}$Code:| | * | /| | r / | | / |y | / | |/θ | - - * - - + - - | x
. . . and that: .$\displaystyle \begin{array}{ccc}r & = & \sqrt{x^2+y^2} \\ \theta & = & \tan^{-1}\!\left(\frac{y}{x}\right)\end{array}$
Then use those conversion formulas . . .
$\displaystyle \begin{Bmatrix}x & = & 3\cos40^o & \approx & 2.3 \\
y & = & 3\sin40^o & \approx & 1.9\end{Bmatrix}\qquad (3,\,40^o)\;\Rightarrow\;2.3i + 1.9j$
$\displaystyle \begin{Bmatrix}r & = & \sqrt{2^2+5^2} & = & \sqrt{29} \\
\theta & = & \tan^{-1}\!\left(\frac{5}{2}\right) & \approx & 68.2^o\end{Bmatrix}\qquad2i + 5j\;\Rightarrow\;\left(\sqrt{29},\,68.2^o\right)
$
use the dot product formula
Recall: If $\displaystyle \bold {a,b}$ are two vectors, then $\displaystyle \boxed { \bold {a} \cdot \bold {b} = |\bold {a}||\bold {b}|\cos \theta }$ where $\displaystyle \theta$ is the angle between them.
you have enough information to solve that formula for $\displaystyle \theta$
do you remember how to take the dot product?
Hello, sir nerdalot!
I thought about trying to use the cosine/sine rule . . . . Good thinking!
but found that there wasn't enough information to do this.
. . Well, there is . . . but it takes some Olympic-level gymnastics.
Given two vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$, let $\displaystyle \theta$ be the angle between them.We have: .$\displaystyle \vec{u} \,= \,\overrightarrow{OP},\;\;\vec{v} \,=\, \overrightarrow{OQ},\;\;\theta \,= \,\angle POQ$Code:Q * * \ v * \ * \ w * \ * θ \ * * * * * * * * * * O u P
. . If $\displaystyle \vec{w} \,= \,\overrightarrow{PQ}$, then: .$\displaystyle \vec{w} \:= \:\vec{u} - \vec{v}$
Using the Law of Cosines, we have: .$\displaystyle |w|^2 \;=\;|u|^2 + |v|^2 - 2|u||v|\cos\theta$ .[1]
A small detour:
. . $\displaystyle |\vec{w}|^2 \;=\;|\vec{u} - \vec{v}|^2 \;=\;(\vec{u}-\vec{v})\cdot(\vec{u}-\vec{v}) \;=\;\vec{u}\cdot\vec{u} - \vec{u}\cdot\vec{v} - \vec{v}\cdot\vec{u} + \vec{v}\cdot\vec{v} $
. . $\displaystyle |w|^2\;=\;|\vec{u}|^2 - 2(\vec{u}\cdot\vec{v}) + |\vec{v}|^2$ .[2]
Substitute [2] into [1]: .$\displaystyle |\vec{u}|^2 - 2(\vec{u}\cdot\vec{v}) + |\vec{v}|^2 \;=\;|\vec{u}|^2 + |\vec{v}|^2 - 2|\vec{u}||\vec{v}|\cos\theta$
This simplifies to: .$\displaystyle 2|\vec{u}||\vec{v}|\cos\theta \:=\:2(\vec{u}\cdot\vec{v})\quad\Rightarrow\quad\b oxed{\cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}| }} $ . . . . There!