# Why cross product gives area of parallelogram formed by two vectors

Recently we were introduced to the concept of vectors in our class, and we learnt about dot products and cross products.I do know that $$a\times b$$ yields the area of the parallelogram formed by the two vectors. What I struggle to understand is why taking the determinant leads to area of the parallelogram, i.e. how does the following: $$\begin{vmatrix} i & j & k \\ a_{x}&a_{y}&a_{z} \\ b_{x}&b_{y} &b_{z} \\ \end{vmatrix}$$ yields a vector which is perpendicular to both $$a$$ and $$b$$, and also has a magnitude$$=ab sin(\theta)$$? How are they connected ?

• The notation you use for that magnitude reminds one of Absinthe which is a French liquor and was according to this answer the main driver behind the invention of differential forms. Jun 21 at 10:57
• This should take you pretty far: \begin{align} \|a\|^2\|b\|^2&=\|a\|^2\|b\|^2\cos^2\theta+\|a\|^2\|b\|^2\sin^2\theta\\ &=(a\cdot b)^2+\|a\times b\|^2 \end{align} For a geometric proof see this answer. Jun 21 at 11:16
• How did your teacher define the cross-product $a\times b$? Jun 21 at 15:34
• @DavidQuinn He simply told us that cross product of two vectors is defined as the area of the parallelogram formed by the two vectors, following which he told us the determinant method to calculate the cross product but without any explanation of how it is derived or proof. Our textbook too skips over this part, I also couldn't find any such proof or derivation on the internet. Jun 21 at 18:28
• The usual definition is $\underline{a}\times\underline{b}=|\underline{a}||\underline{b}|\sin\theta \underline{\hat{n}}$, where $\theta$ is the angle between the vectors, and $\underline{\hat{n}}$ is the unit vector perpendicular to both vectors following the right-hand rule. Jun 21 at 19:41

After reading through others' responses and spending some more time on the problem I have come up with an explanation of my own for this question which I posted and am leaving it as answer over here for public knowledge and introspection.

The determinant method of finding the cross product is basically an abuse of the notation, as explained in this answer, also cited earlier in one of the replies. Ideally a determinant like this should give the volume of the parallelpiped formed by the vector $$a,b$$ and the vector $$\hat{i}+\hat{j}+\hat{k}.$$
Now,generally when we calculate the determinant given in the question, which is: $$\begin{vmatrix} i & j & k \\ a_{x}&a_{y}&a_{z} \\ b_{x}&b_{y} &b_{z} \\ \end{vmatrix}$$ we derive a new vector $$c$$: $$c=(a_{y}b_{z}-a_{z}b_{y})\hat{i}+(a_{z}b_{x}-a_{x}b_{z})\hat{j}+(a_{x}b_{y}-a_{y}b_{x})\hat{k}$$ Upon inspecting the components of the newly found vector one can see that the coefficients of the unit vectors $$\hat{i},\hat{j},\hat{k}$$ are nothing but the area of projections of the parallelogram formed by the vectors $$a$$ and $$b$$ upon the $$y-z, z-x, x-y$$ planes, respectively. When we calculate the magnitude of this vector, we perform the following calculation: $$|c|=\sqrt{{(a_{y}b_{z}-a_{z}b_{y})}^{2}+{(a_{z}b_{x}-a_{x}b_{z})}^{2}+{(a_{x}b_{y}-a_{y}b_{x})}^{2}}$$ Let the areas of these projections be A,B,C respectively.What now needs to be proven is: $$|c|^2=A^2+B^2+C^2$$ But I will be showing this for a general case instead, that is for any closed curve in 3D-space which lies on a single plane and has area $$T$$; $$T^2=(T_{xy})^2+(T_{yz})^2+(T_{zx})^2$$ Where $$(T_{xy}),(T_{yz}),(T_{zx})$$ are its projection upon respective coordinate planes.

First, lets consider two planes $$P_{1},P_{2}$$ inclined at some angle $$\theta$$ with each other. Consider an area element $$S$$ on $$P_{1}$$, then its projection upon the other plane will have an area of $$S(cos(\theta))$$. This can be understood by same approach we apply for calculating area under curves during integration. We use infinitesimally thin strips parallel to each other under the curve. If we take similar strips in plane $$P_{1}$$ which are perpendicular to the line of intersection of $$P_{1}, P_{2}$$ then their projection on the $$P_{2}$$ will be of length $$cos(\theta)$$ times the original length. Therefore any area element($$=S$$) calculated for any curve in $$P_{1}$$ will have its projection on $$P_{2}$$ of area $$=S(cos(\theta))$$.
This holds true irrespective of the orientation of the closed curve with in the plane $$P_{1}$$. For example for a triangle in $$P_{1}$$ with area $$t$$, its projection will have area $$t(cos(\theta))$$. One can rotate or flip the triangle within $$P_{1}$$ and can even deform it to make another shape(but with same area) but its projection's area in $$P_{2}$$ will remain constant as long as the area of the closed curve in $$P_{1}$$ and the $$\angle \theta$$ remains constant.
Second,now employing the De Gua's theorem for a tri-rectangular tetrahedron: Let the right angled sides have area $$e,f,g$$ and the hypotenuse side have area $$h$$,then it follows $$e^2+f^2+g^2=h^2$$
Notice that $$e,f,g$$ can be considered as projected areas of hypotenuse triangle with area $$h$$ on coordinate planes. Let them be the projections with $$xy,yz,zx$$ coordinate planes respectively. And from previous observations we can also conclude that : $$\frac{e}{h}=cos(\alpha)$$ $$\frac{f}{h}=cos(\beta)$$ $$\frac{g}{h}=cos(\gamma)$$ where $$\alpha,\beta,\gamma$$ are the angles made by the hypotenuse face with the respective coordinate planes. Remember that we have already established that no matter how we deform the original closed curve, as long as its area is constant and the containing plane retains its orientation in the 3D space( that is its angles with the coordinate planes($$\alpha,\beta,\gamma$$) remains same). The area of projection will also remain constant.
What that means is that we can translate the hypotenuse triangle in 3D space and deform it in anyway without opening the curve (and keeping area($$h$$) constant) then e,f,h shall also remain constant. Hence, on generalizing, it follows that for any area element($$T$$) in 3D space and contained in single plane: $$T=(T_{xy})^2+(T_{yz})^2+(T_{zx})^2$$ Where $$(T_{xy})^2,(T_{yz})^2,(T_{zx})^2$$ are its projection upon respective coordinate planes.
And thus we conclude that $$A^2+B^2+C^2$$ and $$|c|$$ indeed equals to the area of the parallelogram formed by vectors $$a,b$$.
And $$\therefore$$ $$|c|=|a||b|(sin(\phi)$$, where $$\phi$$ is the angle between the vectors $$a,b$$.
Coming to the perpendicularity of the vector $$c$$ to the plane containing vectors $$a,b$$. To understand this, imagine a unit vector $$u$$ perpendicular to the plane containing $$a$$ and $$b$$, let this plane be $$P_{ab}$$. Let the angles made by the plane $$P_{ab}$$ and coordinate planes $$x-y,y-z,z-x$$ once again be $$\alpha,\beta,\gamma$$ . Since $$u$$ is perpendicular to $$P_{ab}$$ the angles it makes with the coordinate planes $$x-y,y-z,z-x$$ will be $$\frac{\pi}{2}-\alpha,\frac{\pi}{2}-\beta,\frac{\pi}{2}-\gamma$$ And hence the angles it subtends with coordinate axes $$X,Y,Z$$ will be $$\beta,\gamma,\alpha$$ and hence the unit vector $$u$$ will have $$cos(\beta), cos(\gamma),cos(\alpha)$$ along them.Here, just to be clear, $$\beta,\gamma,\alpha$$ are angles of $$u$$ with axes$$X,Y,Z$$ and the angles of $$u$$ with coordinate planes $$y-z,z-x,x-y$$, respectively. Remember that $$c=(a_{y}b_{z}-a_{z}b_{y})\hat{i}+(a_{z}b_{x}-a_{x}b_{z})\hat{j}+(a_{x}b_{y}-a_{y}b_{x})\hat{k}$$ has coefficients as area of projection of parallelogram formed by $$a,b$$ on $$y-z,z-x,x-y$$ for unit vectors $$\hat{i}, \hat{j}, \hat{k}$$ respectively. And we have already established that projected areas are $$cos(\theta)$$ times the original area, $$\theta$$ being the angle between the two planes.
Therefore vector $$c$$ also has unit vector same as vector $$u$$, i.e. $$cos(\beta),cos(\alpha),cos(\gamma)$$. Therefore vector $$c$$ is perpendicular to $$P_{ab}$$.