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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 $\vec{v}$.

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 it 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?.

J. W. Tanner
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Chris Steinbeck Bell
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2 Answers2

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The cross product of two vectors is a vector perpendicular to both of them, and its magnitude is the area of a parallelogram with the vectors for sides. It can be computed for $\vec{u}=\hat{j}+\sqrt{3}\hat{k}$ and $\vec{v}=\sqrt{3}\hat{j}+2\hat{k}$ as follows:

$$\vec u\times\vec v=\begin{vmatrix}\hat i&&\hat j&&\hat k\\0&&1&&\sqrt3\\0&&\sqrt3&&2\end{vmatrix}.$$

Here is additional information about the cross product, added per OP's request (see comments):

The cross product of two vectors will yield a vector perpendicular to both, whereas if two vectors are perpendicular then their dot product will be $0$; so, for example, $\vec a ⋅(\vec a ×\vec b)=0$.

The dot product is commutative, but the cross product is anti-commutative: $\,\vec b ×\vec a =−(\vec a ×\vec b ) $.

See this question for explanations why the magnitude of $\vec a\times\vec b$ gives the area of the parallelogram.

J. W. Tanner
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  • Let me know if you don't understand determinant notation – J. W. Tanner Mar 18 '20 at 00:05
  • Yes I inderstand the determinant notation. I just don't know how to prove that the magnitude of the cross product is equal to the area of the paralellogram. Does it exist an algebraic proof or some sort of it?. – Chris Steinbeck Bell Mar 18 '20 at 00:11
  • So If I want a vector which to be perpendicular to another, I just have to do a cross product and equate to zero?. Am I right with this?. Let's suppose I have $\vec{v}=\left \langle 1,1 \right \rangle$ and I need $\vec{w}$ to be perpendicular to it $\left \langle a,b \right \rangle$ then $\vec{v} \times \vec{w} = 0$ ? Or is it in reverse?. Or there is insuficient information to determine such vector?. – Chris Steinbeck Bell Mar 18 '20 at 00:15
  • If $\vec v\perp \vec w$ then $\vec v\cdot \vec w=0$; $(1,1,0)\times(a,b,0)$ is perpendicular to $(1,1,0)$ and $(a,b,0)$; the usual cross product definition works for $3$-dimensional vectors; proving that the magnitude of the cross product defined with determinant notation is equal to the area of the parallelogram can be done, but is a different question – J. W. Tanner Mar 18 '20 at 00:30
  • Wait! is it the dot product or is it the cross product?. Sorry it seems obvious now that a cross product will always yield a vector perpendicular to both. I was not seeing a three dimentional space. But since is a dot product it will not matter which goes first as it is conmutative right?. – Chris Steinbeck Bell Mar 18 '20 at 00:40
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    Yes, a cross product will yield a vector perpendicular to both, whereas if two vectors are perpendicular then their dot product will be $0$; so, for example $\vec a\cdot( \vec a\times \vec b)=0$; dot product is commutative but cross product is anti-commutative: $\vec b\times \vec a=-(\vec a\times\vec b)$ – J. W. Tanner Mar 18 '20 at 00:44
  • I voice my opinion that you should add your last comment to be part of you answer because that made me understand better and a link to some sort of reference on where's the proof that the area of the paralellogram is equal in magnitude to the cross product of vectors involved. Can you do that please? :). – Chris Steinbeck Bell Mar 18 '20 at 00:50
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$u \times v=-i$. The area of the parallogram is $||u \times v||=||-i||=1.$A vector perpendicular to $$u \text { and } v \text { is } c(u \times v)=-ci$$ whose magnitude is $||-ci||=|c|.$ Thus $|c|=1$, so $c= \pm 1$ and the required vector is $\pm i$.

P. Lawrence
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