User:Ujin-X/sandbox

See how a similar example is solved using the angular and inverse vectors.

Example: Proposed changes to the cross product. When at the cross product of two rectilinear vectors, a angular vector is created $$\quad$$ $d&#785;$ $$=\bar{b}\times\bar{c}$$. This vector is in the plane formed by rectilinear vectors. Its magnitude is equal to the area of some figure in this plane. Suppose we are given two rectilinear vectors $$\bar{b}(10, 0, 0), \bar{c}(0, 10, 0) $$, then $d&#785;$$$=\bar{b}\times\bar{c}=(b_{x}c_{y}-b_{y}c_{x})$$$l&#785;$$$+(b_{y}c_{z}-b_{z}c_{y})$$$m&#785;$$$+(b_{z}c_{x}-b_{x}c_{z})$$$n&#785;$$$=d_{xy}$$$l&#785;$$$+d_{yz}$$$m&#785;$$$+d_{zx}$$$n&#785;$$$ \quad$$, where $l&#785;$,$m&#785;$,$n&#785;$ - the basic angular vectors $d&#785;$$$ =(10\cdot 10-0\cdot 0)$$$l&#785;$$$+(0\cdot 0-0\cdot 10)$$$m&#785;$$$+(0\cdot 0-10\cdot 0)$$$n&#785;$$$=100$$$l&#785;$$$+0$$$m&#785;$$$+0$$$n&#785;$$$ $$ $$ d_{xy}=100, d_{yz}=0, d_{zx}=0 $$, where $$d_{xy},d_{yz},d_{zx}$$ - the projections of the angular vector onto the coordinate planes. That is, the angular vector $d&#785;$$$(100, 0, 0) $$ is in the plane OXY

Now we find the rectilinear vector $$\bar{c}\quad$$ again and compare it with the original vector. For this we use the inverse vector. Thanks to it vectors can be transferred through the sign =, as in arithmetic. But the choice of a sequence of vectors in a cross product changes the results of calculations, and is always chosen individually. For the solution, we need the cross products of the basis vectors $$\bar{i}\times$$$l&#785;$$$=\bar{j},\quad\bar{j}\times$$$m&#785;$$$=\bar{k},\quad\bar{k}\times$$$n&#785;$$$=\bar{i},\quad\bar{j}\times$$$l&#785;$$$=-\bar{i},\quad\bar{k}\times$$$m&#785;$$$=-\bar{j},\quad\bar{i}\times$$$n&#785;$$$=-\bar{k}\quad$$,then $$\bar{c}=\dfrac{1}{\bar{b}}\times$$$d&#785;$$$=\bar{b'}\times$$$d&#785;$$$=(b'_{x}\bar{i}+b'_{y}\bar{j}+b'_{z}\bar{k})\times(d_{xy}$$$l&#785;$$$+d_{yz}$$$m&#785;$$$+d_{zx}$$$n&#785;$$$)=$$ $$=(b'_{z}d_{zx}-b'_{y}d_{xy})\bar{i}+(b'_{x}d_{xy}-b'_{z}d_{yz})\bar{j}+(b'_{y}d_{yz}-b'_{x}d_{zx})\bar{k}=$$ Further, we use the projections of the inverse rectilinear vector $$\quad b'_{x}=\dfrac{b_{x}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}};\quad b'_{y}=\dfrac{b_{y}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}\quad b'_{z}=\dfrac{b_{z}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}$$ $$=\left(\dfrac{b_{z}d_{zx}-b_{y}d_{xy}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}\right)\bar{i}+\left(\dfrac{b_{x}d_{xy}-b_{z}d_{yz}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}\right)\bar{j}+\left(\dfrac{b_{y}d_{yz}-b_{x}d_{zx}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}\right)\bar{k}=c_{x}\bar{i}+c_{y}\bar{j}+c_{z}\bar{k}$$ $$c_{x}=\dfrac{b_{z}d_{zx}-b_{y}d_{xy}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}=\dfrac{0\cdot 0-0\cdot 100}{10^{2}+0+0}=0,$$ $$ c_{y}=\dfrac{b_{x}d_{xy}-b_{z}d_{yz}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}=\dfrac{10\cdot 100-0\cdot 0}{10^{2}+0+0}=10,$$ $$ c_{z}=\dfrac{b_{y}d_{yz}-b_{x}d_{zx}}{b_{x}^{2}+b_{y}^{2}+b_{z}^{2}}=\dfrac{0\cdot 0-10\cdot 0}{10^{2}+0+0}=0$$

that is there is a vector $$ \vec{c}(0, 10, 0)$$, and is equal to the original vector. Now imagine that the angular vector $d&#785;$ is the torque. A rectilinear vector $c&#772;$ is a force, and a rectilinear vector $b&#772;$, this distance. And this example shows that we can find in the coordinate-vector form a torque from the force and distance, which is in the plane, where it should be. And we can solve the inverse problem, find the force from the torque, again in a coordinate-vector form. I emphasize that we could not find the last task before. And she decided because of the inverse vector and its projections.

Inverse vector


An inverse rectilinear vector $a&#772;$' is a vector which is co-directed (in the same direction as) a vector $a&#772;$ and differs from it in magnitude according to: $$\quad |$$$a&#772;$'$$|=\dfrac{1}{|\text{ā}|}$$ Projections on the coordinate axes of inverse rectilinear vectors are equal according to: $$\quad a'_{x}=\dfrac{a_{x}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}};\quad a'_{y}=\dfrac{a_{y}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}\quad a'_{z}=\dfrac{a_{z}}{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}$$

An inverse angular vector $a&#785;$' is similarly a vector which is co-directed with a vector $a&#785;$ and differs from it in magnitude: $$\quad |$$$a&#785;$'$$|=\dfrac{1}{|\text{ȃ}|}$$ Projections on the coordinate planes of inverse angular vector are equal according to: $$\quad a'_{xy}=\dfrac{a_{xy}}{a_{xy}^{2}+a_{yz}^{2}+a_{zx}^{2}};\quad a'_{yz}=\dfrac{a_{yz}}{a_{xy}^{2}+a_{yz}^{2}+a_{zx}^{2}}\quad a'_{zx}=\dfrac{a_{zx}}{a_{xy}^{2}+a_{yz}^{2}+a_{zx}^{2}}$$

Inverse vectors are used in mathematics to perform vector division operations.