Area of Parallelogram
- Thread starter Lime
- Start date
Hello, Lime!
skeeter is absolutely correct . . .
First, find: \(\displaystyle \:\vec{u}\,\times\,\vec{v}\;=\;\begin{vmatrix}i & j & k \\ -1 & 1 & 0 \\ 2 & 3 & -1\end{vmatrix}\;=\;-i\,-\,j\,-\,5k \;=\;\langle-1,-1,-5\rangle\)
Then: \(\displaystyle \:\text{Area }\:=\:\left|\vec{u}\,\times\,\vec{v}\right| \;=\;\sqrt{(-1)^2\,+\,(-1)^2\,+\,(-5)^2} \;=\;\sqrt{27} \;=\;3\sqrt{3}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you are unfamiliar with that area formula,
. . you could have found the area anyway.
The area of a parallelogram is: \(\displaystyle \,\text{base }\times\text{ height.}\)
The base is \(\displaystyle |u|\) . . . The height is \(\displaystyle |v|\sin\theta\)
. . Hence, the area is: \(\displaystyle \:A \:=\:|u||v|\sin\theta\:\) [1]
The angle \(\displaystyle \theta\)between two vectors \(\displaystyle \vec{u}\) and \(\displaystyle \vec{v}\) is given by: \(\displaystyle \L\:\cos\theta \;=\;\frac{|\vec{u}\cdot\vec{v}|}{|\vec{u}||\vec{v}|}\)
We have: \(\displaystyle \,\left|\vec{u}\cdot\vec{v}\right|\:=\:\left|\langle-1,1,0\rangle\cdot\langle2,3,-1\rangle\right| \:=\:\left|-2\,+\,3\,+\,0\right| \:=\:1\)
. . and: \(\displaystyle \:|\vec{u}| \:=\:\sqrt{(-1)^2\,+\,1^2\,+\,0^2} \:=\:\sqrt{2}\,\) and \(\displaystyle \,|\vec{v}| \:=\:\sqrt{2^2\,+\,3^2\,+\,(-1)^2} \:=\:\sqrt{14}\)
Hence: \(\displaystyle \:\cos\theta \:=\:\frac{1}{\sqrt{28}} \;\;\Rightarrow\;\;\sin\theta \:=\:\frac{3\sqrt{3}}{\sqrt{28}}\)
Substitute into [1]: \(\displaystyle \L\:A \:=\:|\vec{u}||\vec{v}|\sin\theta \:=\
\sqrt{2})(\sqrt{14})\left(\frac{3\sqrt{3}}{\sqrt{28}\right) \:=\:\fbox{3\sqrt{3}}\)
. . . . . see?
skeeter is absolutely correct . . .
First, find: \(\displaystyle \:\vec{u}\,\times\,\vec{v}\;=\;\begin{vmatrix}i & j & k \\ -1 & 1 & 0 \\ 2 & 3 & -1\end{vmatrix}\;=\;-i\,-\,j\,-\,5k \;=\;\langle-1,-1,-5\rangle\)
Then: \(\displaystyle \:\text{Area }\:=\:\left|\vec{u}\,\times\,\vec{v}\right| \;=\;\sqrt{(-1)^2\,+\,(-1)^2\,+\,(-5)^2} \;=\;\sqrt{27} \;=\;3\sqrt{3}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you are unfamiliar with that area formula,
. . you could have found the area anyway.
Code:
*- - - - - - - - - - *
/: /
v / : /
/ : v·sinθ /
/ : /
/θ : /
* - - - - - - - - - - *
uThe area of a parallelogram is: \(\displaystyle \,\text{base }\times\text{ height.}\)
The base is \(\displaystyle |u|\) . . . The height is \(\displaystyle |v|\sin\theta\)
. . Hence, the area is: \(\displaystyle \:A \:=\:|u||v|\sin\theta\:\) [1]
The angle \(\displaystyle \theta\)between two vectors \(\displaystyle \vec{u}\) and \(\displaystyle \vec{v}\) is given by: \(\displaystyle \L\:\cos\theta \;=\;\frac{|\vec{u}\cdot\vec{v}|}{|\vec{u}||\vec{v}|}\)
We have: \(\displaystyle \,\left|\vec{u}\cdot\vec{v}\right|\:=\:\left|\langle-1,1,0\rangle\cdot\langle2,3,-1\rangle\right| \:=\:\left|-2\,+\,3\,+\,0\right| \:=\:1\)
. . and: \(\displaystyle \:|\vec{u}| \:=\:\sqrt{(-1)^2\,+\,1^2\,+\,0^2} \:=\:\sqrt{2}\,\) and \(\displaystyle \,|\vec{v}| \:=\:\sqrt{2^2\,+\,3^2\,+\,(-1)^2} \:=\:\sqrt{14}\)
Hence: \(\displaystyle \:\cos\theta \:=\:\frac{1}{\sqrt{28}} \;\;\Rightarrow\;\;\sin\theta \:=\:\frac{3\sqrt{3}}{\sqrt{28}}\)
Substitute into [1]: \(\displaystyle \L\:A \:=\:|\vec{u}||\vec{v}|\sin\theta \:=\
\sqrt{2})(\sqrt{14})\left(\frac{3\sqrt{3}}{\sqrt{28}\right) \:=\:\fbox{3\sqrt{3}}\). . . . . see?