Area of Parallelogram

[Lime]

Find the area of the parallelogram determined by the vectors u = (-1, 1, 0) and v = (2, 3, -1).

 

[skeeter]

the area of the parallelogram will be the magnitude of the cross product of vectors u and v.

 

[soroban]

Hello, Lime!

skeeter is absolutely correct . . .

Find the area of the parallelogram determined
by the vectors: $\displaystyle \,\vec{u} \:= \:\langle-1,1,0\rangle\,$ and $\displaystyle \,\vec{v} \:= \:\langle2,3,-1\rangle$

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: $\displaystyle \:\text{Area }\:=\:\left|\vec{u}\,\times\,\vec{v}\right| \;=\;\sqrt{(-1)^2\,+\,(-1)^2\,+\,(-5)^2} \;=\;\sqrt{27} \;=\;3\sqrt{3}$

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If you are unfamiliar with that area formula,
. . you could have found the area anyway.

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                  u

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?