Finding area of a parallelogram

asked 2014-06-25 12:27:36 -0600

Tiffany
656 ●3 ●18 ●34

updated 2014-06-30 15:58:34 -0600

Justin
1009 ●2 ●20 ●38

So I've been working on problem 7 in 12.4. Following the book on page 315 about parallelograms.

"Given two vectors, we can put them tail to tail and form a parallelogram...The height of the parallelogram, h, is |A|$ \sin(\theta)$ , and the base is |B|, so the area of the parallelogram is |A||B|$ \sin(\theta)$, exactly the magnitude of |A x B|."

The problem states "7. Find the area of the parallelogram with vertices (0, 0), (1, 2), (3, 7), and (2, 5)."

When I sketched out the parallelogram with the given points, it looked like the base of the parallelogram would be the vector from the origin to (1,2) (B), and the top vector would be (2,5)(A). (I'm not sure if this is where my mistake was, or if it was in trying to find $\theta$.)

Using those two points as the vectors, I ended up with |A| = $\sqrt{29}$ and |B| =$ \sqrt{5}. $

I then used the dot product |A| * |B| = |A||B|$\cos$($\theta$) to attempt to find theta. Plugging everything in I ended up with $$ \theta = \cos^{-1}(\frac {12} {\sqrt{29} \sqrt{5}}) $$

Plugging that into the area equation from the book, I ended up with..

$$ area = \sqrt{29}\sqrt{5}(\sin(\cos^{-1}(\frac {12} {\sqrt{29}\sqrt{5}})))$$

Which unless I did wrong in my calculator gave me 0. But the back of the book showed that the answer should have been 1. So I'm not exactly sure where I went wrong, or if I did the whole problem wrong. So if anybody could give some input that would be fantastic!

Thanks!

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answered 2014-06-25 17:09:15 -0600

Justin
1009 ●2 ●20 ●38

This is a good problem since it shows that there is almost always more than one way to come up with a solution. Instead of dealing with sine, cosine and theta, I just calculated what $A \times B$ actually was and calculated its magnitude. This way, a calculator is not needed at all, which may be helpful for a situation where calculators are not allowed. Here is how I did this problem (note that I kept the $z$-component of the vectors as $0$ for simplicity of calculation):

$$ |A \times B| = \left| \begin{array}{ccc} \vec \imath & \vec \jmath & \vec k \newline 1 & 2 & 0 \newline 2 & 5 & 0 \newline \end{array} \right| = \vec \imath (0 - 0) - \vec \jmath (0 - 0) + \vec k(5 - 4) = \langle 0, 0, 1 \rangle $$

The magnitude of $\langle 0, 0, 1 \rangle$ equals $1$, no calculator needed!

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answered 2014-06-25 13:04:01 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

I followed your logic and came up with the same figures and answer for the area. (Great job, by the way!)

However, when I entered it into my calculator, I did in fact come up with 1. My suggestion would be to go back and try the last calculation again taking care with how you enter the radicals and parentheses, etc. into your calculator.

I often take extra steps because I am so bad to make entry errors.

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Finding area of a parallelogram

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