I was shocked to find that the area of parallelogram in two dimensions can be found by cross multiplying two adjacent vectors.   I understand that in three dimensions the area of parallelogram is the cross product of two adjacent vectors $$|a \times b|$$ but no text books have explained why in two dimensions cross product of two adjacent vectors are the area of a parallogram. Could someone explain?
122758
I tried looking it up but many websites just state it without proof and without intuition.  I'm hoping to learn it a little bit better so that I don't forget how to compute the Jacobian when working with surface integrals where the divergence theorem is not applicable.  If you have a good online reference instead, please feel free to provide it :-)  Thanks,
76556
My professor told me recently that Area is a vector. A Google search gave me the following definition for a vector:  Noun: A quantity having direction as well as magnitude, esp. as determining the position of one point in space relative to another.  My question is - what is the direction of area? I can relate to the fact that velocity is a vector. The velocity of a moving motorbike for example, has a definite direction as well as a definite magnitude assuming that the bike is moving in a straight line &amp; not accelerating. My friend gave me this explanation for the direction of Area vector. Consider a rectangular plane in space. He argued that the orientation of the plane in space can only be described by considering area as a vector &amp; not a scalar. I still wasn't convinced. Suppose the plane was placed such that its faces were perpendicular to the directions, North &amp; South for example. Now the orientation of the plane is the same irrespective whether the so called vector points to north or to the south. Further what is the direction of a sphere's area? Does considering area as a vector have any real significance? Please explain. Thanks in advance.
111342
I was given this question by a friend and after working tirelessly on it I have not come up with anything substantial. I was hoping someone in the community could provide a pointer or possibly a solution. The person who gave me the question says he has the answer to it, but I am more curious about the working.  Here is the question:     Three points $(B, C, D)$ lie on parabola  $y = x^2$ and a fourth point $A$ is placed such that $ABCD$ forms a square. Find the minimal area of $ABCD$.   So far my working out has involved a collection of distance/mid-point formula's and I haven't even reached the point where I differentiate to minimize the area.  Any pointers would be welcome.   Edit  Apparently the correct answer is $\sqrt{2}$ units squared.
19597
I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.  If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a parallelogram.  It is assignment from linear algebra class. I wasn't sure if I had to like use a matrix or something.
78749
I read  a while ago, and while thinking about $\nabla$, I realized something. Since the cross product can be written as a determinant, in higher dimensions we require extra vector inputs. IIRC it's called the "wedge product" in higher dimensions.  Alright, how does this work when we generalize Maxwell to higher dimensions? Curl can be written (abuse of notation, yes) as a cross product with $\nabla$. But, to generalise it to higher dimensions, we need multiple inputs. We need something like $\nabla_4(\mathbf{B_1},\mathbf{B_2})$  in four dimensions, and so on. So we have two ways to get out of this: We can either use a different way to write curl in multiple dimensions (The  has stuff on this which I don't understand), or there are more than one $E$ and $B$ fields in higher dimensions.  So which is it? How are Maxwell's laws generalised to higher dimensions?  Just a note: I never understood the linked answer after the first sentence (didn't know enough), so if there's something obvious there that answers this question, I missed it. I know nothing of higher-dimensional analysis, so if complex notation is going to be unavoidable for the multiple dimensions, I'd be fine if you showed me what happens in four dimensions.
78307
As far as I've seen, there are only two types of vector multiplication defined, dot product/inner product which is defined in a vector space of any dimension, even infinite, and cross product, which is only defined in three dimensions. Are there more definitions of vector products?
11856
I get the graphic explanation, i.e. that the area $dA$ of the sector's increment can be looked upon as a polar "rectangle" as $dr$ and $d\theta$ are infinitesimal, but how do you prove this rigorously?    1. Geometrically, the exact area would be $$\frac{(r+dr)^2d\theta}{2} - \frac{r^2d\theta}{2}$$$$= (r + \frac{dr}{2}) dr d\theta  $$$$= r dr d\theta + \frac{dr^2 d\theta}{2}.$$  How do we get rid of $\frac{dr^2 d\theta}{2}$? Is it too insignificant in value compared to $r dr d\theta$ so that it vanishes? If so, is it the reason it gets ignored? And what's more, geometrically speaking, why would the two areas - $dx dy$ and $r dr d\theta$, be equal?  2. Approaching it algebraically, setting $$x = r\sin\theta$$$$y = r \cos\theta$$  gives $$\frac{dx}{d\theta} = r\cos\theta,  dx = r\cos{\theta} d\theta$$$$ \frac{dy}{dr} = \cos\theta,  dy = \cos\theta dr. $$  Multiplying both equations, side by side, gives $$dxdy = r\cos^2\theta dr d\theta.$$  Again I get an extra term, which is $\cos^2 \theta$. In both cases I am unable to derive that $dxdy = rdrd\theta$.  What do I do wrong in my reasoning? It all makes me think that I'm getting something essential terribly wrong.
71750
Note that a product   may be parenthesized in two different ways:   and . Similarly, there are several different ways to parenthesize . Two such ways are  and . Let  be the number of different ways to parenthesize the product . Show that if , then      for all integers .  Keep in mind that the sequence  is the same as the sequence of Catalan numbers.   I am entirely lost. Please can someone offer some hint or direction of where to begin? I humbly thank you so very much.
46997
I know how to CALCULATE dot product, but still don't understand MEANING of it.  I mean, If I have to explain vector addition I would say that "Sum of two vectors give you the third vector that has the same effect as other two combined"  But I don't know how to explain dot product. For example we have two vectors: v=(3,1) and w=(2,4). Dot product of these two is 3*2 + 1*4 = 10. What does this result (10) tell us?
29219
Suppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.  My question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).
19588
I was wondering if anyone could give me the intuition behind the cross product of two vectors $\textbf{a}$ and $\textbf{b}$. Why does their cross product $\textbf{n} = \textbf{a} \times \textbf{b}$  give me a vector which is perpendicular to a plane?  I know I can just check this by using dot product but I'm not totally satisfied with "it just works" answer =)  Thank you for any help! =)
62869
How do I find a vector perpendicular to a vector like this: $$3\mathbf{i}+4\mathbf{j}-2\mathbf{k}?$$ Could anyone explain this to me, please?  I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components...  When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks.
22172
Prove that: area of triangle $S_{ABC}=p \cdot R$, where p is half perimeter of triangle $DEF$. $AD,CF, BE$ are altitudes. Is there a simple way?
47958
What's the analogue to spherical coordinates in $n$-dimensions? For example, for $n=2$ the analogue are polar coordinates $r,\theta$, which are related to the Cartesian coordinates $x_1,x_2$ by  $$x_1=r \cos  \theta$$ $$x_2=r \sin  \theta$$  For $n=3$, the analogue would be the ordinary spherical coordinates $r,\theta ,\varphi$, related to the Cartesian coordinates $x_1,x_2,x_3$ by  $$x_1=r \sin  \theta  \cos  \varphi$$ $$x_2=r \sin  \theta  \sin  \varphi$$ $$x_3=r \cos  \theta$$  So these are my questions: Is there an analogue, or several, to spherical coordinates in $n$-dimensions for $n&gt;3$? If there are such analogues, what are they and how are they related to the Cartesian coordinates? Thanks.
97089
We know how to make cross product of three dimensional vectors. $$ \vec A \times \vec B = \vec C$$  Where : $ \vec A = (A_i; A_j; A_k)$ $ \vec B = (B_i; B_j; B_k)$ $ \vec C = (C_i; C_j; C_k)$  $C_i = \left|\begin{matrix}A_j&amp;A_k\\B_j&amp;B_k\end{matrix}\right|$ $C_j = \left|\begin{matrix}A_k&amp;A_i\\B_k&amp;B_i\end{matrix}\right|$ $C_k = \left|\begin{matrix}A_i&amp;A_j\\B_i&amp;B_j\end{matrix}\right|$  But what about if we have four dimensional vectors? Is it possible to make cross product of four dimensional vectors? If it  is possible, then tell me when it can be possible?       Let say we have two vectors: $ \vec A = (A_i; A_j; A_k; A_l)$ $ \vec B = (B_i; B_j; B_k; B_l)$   Then how to compute a cross product of this two vectors? Will it again vector? $$ \vec A \times \vec B = \vec C$$  $ \vec C = (C_i; C_j; C_k; C_l)$ Then how to compute those coordinates?  We know that only square matrices have a determinant property! In this case it might not be correct if we will wright... $\color{red} {\text { $C_i = \left|\begin{matrix}A_j&amp;A_k&amp;A_l\\B_j&amp;B_k&amp;B_l\end{matrix}\right|$}    C_j = \left|\begin{matrix}A_k&amp;A_i&amp;A_l\\B_k&amp;B_i&amp;B_l\end{matrix}\right| C_k = \left|\begin{matrix}A_i&amp;A_j&amp;A_l\\B_i&amp;B_j&amp;B_l\end{matrix}\right| C_l = \left|\begin{matrix}A_i&amp;A_j&amp;A_k\\B_i&amp;B_j&amp;B_k\end{matrix}\right|}$   So tell me how to solve this problem?
46219
If $\vec{a},\vec{b},\vec{c},\vec{d}$are unit vectors such that $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ and $\vec{a}.\vec{c}=\frac{1}{2}$,then $(A)\vec{a},\vec{b},\vec{c}$ are non-coplanar $(B)\vec{a},\vec{b},\vec{d}$ are non-coplanar $(C)\vec{b},\vec{d}$ are non-parallel. $(D)\vec{a},\vec{d}$ are parallel and $\vec{b}$ and $\vec{c}$ are parallel.    My Attempt: $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ $(\vec{a}.\vec{c})(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$ $\frac{1}{2}(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$    Book's Solution which i did not understand: $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ is possible only when $|\vec{a}\times\vec{b}|=|\vec{c}\times\vec{d}|=0$ and $(\vec{a}\times\vec{b})$ is parallel to $(\vec{c}\times\vec{d})$ And the correct option given is $(C)$.  I could not solve this problem after some efforts.Please help me.Thanks.
15334
Below the parallelogram is obtained from square by stretching the top side while fixing the bottom. Since area of parallelogram is base times height, both square and parallelogram have the same area.    This is true no matter how far I stretch the top side.    In below figure it is easy to see why both areas are same.   But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?