How can I find the magnitude of a vector which is the same as the area of a parallelogram?

Question

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?.

Answer 1

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:

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

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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.

Answer 2

$u \times v=-i$. 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 $\pm i$.