How can I find the magnitude of a vector when it is the same to the area of a paralellogram?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

Find a vector which is perpendicular to the vectors $\vec{u}=\hat{j}+\sqrt{3}\hat{k}$ and $\vec{v}=\sqrt{3}\hat{j}+2\hat{k}$ whose magnitude is equal to the area of the parallelogram which is formed by $\vec{u}$ and $\vec{b}$.

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&-2\hat{j}\\
2.&-\hat{i}\\
3.&3\hat{k}\\
4.&5\hat{i}\\
5.&3\hat{i}\\
\end{array}$

I'm totally lost at this question. What should I do to find the area? Does there exist a formula which can be used to relate it with the fact that a vector is perpendicular to those two? Can someone explain the solution step by step, so I can understand it?
 

topsquark

Math Team
May 2013
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Hint: Note that the cross product of two vectors gives a vector that is perpendicular to both. What does the magnitude of this vector represent?

-Dan
 
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Jun 2017
399
6
Lima, Peru
Hint: Note that the cross product of two vectors gives a vector that is perpendicular to both. What does the magnitude of this vector represent?

-Dan
Would it be that is the area of the paralelogram? How can I prove it? Is it the magnitude equal or is it just proportional? By the way in the problem I forgot to mention that it should had been $\vec{v}$ not $\vec{b}$. Perhaps can you help me with this part? I mean the proof.
 

skeeter

Math Team
Jul 2011
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Jun 2017
399
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Lima, Peru
Nice article, but one aspect which I don't get is. Cross product is equal in magnitude to the area made by the vectors? Or is it proportional to it?

From solving this thing, I'm getting $-\hat{i}$ which would be the vector perpendicular to both. Can you help me to clear out this doubt?
 

skeeter

Math Team
Jul 2011
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Texas
re-read the #2 topic in the link again.
 
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topsquark

Math Team
May 2013
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The Astral plane
Nice article, but one aspect which I don't get is. Cross product is equal in magnitude to the area made by the vectors? Or is it proportional to it?

From solving this thing, I'm getting $-\hat{i}$ which would be the vector perpendicular to both. Can you help me to clear out this doubt?
Another hint: What is the formula for a parallelogram if you know the two different sides? That should guide you.

Yes, the unit vector \(\displaystyle \hat{i}\) is a perpendicular (so is \(\displaystyle - \hat{i}\) ) but you are going to want to look at the magnitude of the cross product as well. The magnitude of the cross product is \(\displaystyle | \vec{a} \times \vec{b} | = | \vec{a} | \cdot | \vec{b} | \cdot \sin( \theta )\), where \(\displaystyle \theta \) is the angle between the two vectors.

-Dan
 
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Jun 2017
399
6
Lima, Peru
Another hint: What is the formula for a parallelogram if you know the two different sides? That should guide you.

Yes, the unit vector \(\displaystyle \hat{i}\) is a perpendicular (so is \(\displaystyle - \hat{i}\) ) but you are going to want to look at the magnitude of the cross product as well. The magnitude of the cross product is \(\displaystyle | \vec{a} \times \vec{b} | = | \vec{a} | \cdot | \vec{b} | \cdot \sin( \theta )\), where \(\displaystyle \theta \) is the angle between the two vectors.

-Dan
Then the magnitude of the cross product is equal to the area between two vectors. But it doesn't make sense that the area is negative? Or should I understand that the vector has a negative sign but the area is just an absolute value from it? Am I getting the right picture here?
 

skipjack

Forum Staff
Dec 2006
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The magnitude of a vector cannot be negative.
 
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