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Let ${v} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}$ and ${w} = \begin{pmatrix} 11 \\ -2 \end{pmatrix}$. Find the area of the parallelogram with vertices ${0}$, ${v}$, ${w}$, and $v + {w}$.

What is the best way to start this?

Thanks

Rohini Sharma
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    Hint: Area of the parallelogram with the vectors $u$ and $v$ as sides is $A = |u \times v|$. – Kaster Aug 09 '15 at 18:13

4 Answers4

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Just use determinant formula - | u x w |.

det(5,-3) x det(11,-2) 

=(5 x -2) - (11 x -3)

=23

P.S. - I am sorry I am new to Maths SE so doesn't know how to edit this stuff. But , I try to help if I can.

Shubham Marathia
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You can directly apply something about determinants.

Alternatively, find the fourth vertex of the parallelogram and then do everything from scratch. Maybe I could post something about working "from scratch" if I were incited enough.

Michael Hardy
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That question is begging to be answered with determinants. The area of the parallelogram formed by your vectors $v$ and $w$ is:

$$A = \left|\det \begin{pmatrix} 5 & 11 \\ -3 & -2 \end{pmatrix} \right| = 23$$

Here's an image I stole from Wikipedia:
user123
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If you don't want to do it with determinants as the other answers suggest, but would rather like a more classical geometrical approach, the area of the parallelogram is twice the area of the triangle $\triangle 0vw$. And that triangle's area can be found using Heron's formula.

Arthur
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