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 June 8th, 2017, 08:07 AM #1 Member   Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0 Vectors again Find the area of parallelogram given by the vectors e=2m-n and f=4m-5n. |m|=|n|=1 and the angle between m and n is pi/4. I really struggle with this one and any help is appreciated
 June 9th, 2017, 03:44 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 The area of a parallelogram is, of course, given by "bh" where b is the length of one side (the "base" which we can take to be e) and h the height. If $\displaystyle \theta$ is the angle between the two sides, then $\displaystyle h= \tan(\theta)|e|$ so that the area is $\displaystyle \tan(\theta)|e||f|$. Here, you are given that |e|= |f|= 1 and $\displaystyle \theta$ is $\displaystyle \pi/4$ radians. Last edited by skipjack; June 9th, 2017 at 06:14 AM.
June 9th, 2017, 05:50 AM   #3
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Quote:
 Originally Posted by Country Boy The area of a parallelogram is, of course, given by "bh" where b is the length of one side (the "base" which we can take to be e) and h the height. If $\displaystyle \theta$ is the angle between the two sides, then $\displaystyle h= \tan(\theta)|e|$ so that the area is $\displaystyle \tan(\theta)|e||f|$. Here, you are given that |e|= |f|= 1 and $\displaystyle \theta$ is $\displaystyle \pi/4$ radians.
Yeah, but when give vectors the area is |exf| isn't it?

Last edited by skipjack; June 9th, 2017 at 06:15 AM.

 June 9th, 2017, 06:21 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 Yes. Country Boy slipped up. What's stopping you from evaluating |e × f|?
June 9th, 2017, 06:45 AM   #5
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Quote:
 Originally Posted by skipjack Yes. Country Boy slipped up. What's stopping you from evaluating |e × f|?
I did this, but the textbook says I'm wrong.
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 June 9th, 2017, 09:28 AM #6 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 What does the textbook give as the answer?
June 9th, 2017, 09:32 AM   #7
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Quote:
 Originally Posted by skipjack What does the textbook give as the answer?
3sqrt(2))/2

Last edited by skipjack; June 9th, 2017 at 07:55 PM.

June 9th, 2017, 10:06 AM   #8
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Quote:
 |m|=|n|=1 and the angle between m and n is pi/4
could this statement mean $m$ and $n$ are unit vectors that are not orthogonal?

if so ...

$e = \left(2-\dfrac{1}{\sqrt{2}}\right)i - \left(\dfrac{1}{\sqrt{2}}\right)j$

$f = \left(4-\dfrac{5}{\sqrt{2}}\right)i - \left(\dfrac{5}{\sqrt{2}}\right)j$

however, this makes $|e \times f| = 3\sqrt{2}$

?

 June 9th, 2017, 08:09 PM #9 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 The answer depends on the meaning of "given by the vectors" in the problem. If the specified vectors correspond to the sides of the parallelogram, the area is 3√2. If the vectors correspond to the diagonals of the parallelogram, the area is 3/√2.
June 10th, 2017, 12:12 AM   #10
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Quote:
 Originally Posted by skipjack The answer depends on the meaning of "given by the vectors" in the problem. If the specified vectors correspond to the sides of the parallelogram, the area is 3√2. If the vectors correspond to the diagonals of the parallelogram, the area is 3/√2.
It's the sides. Probably it's just mistake in the textbook Thank you.

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