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For the scenario described as the follows:

Two vectors $\vec{s}$ and $\vec{t}$ lie in the xy plane. Their directions are, respectively, a° and b° measured counterclockwise from the positive x axis. Find the magnitude and the direction of $\vec{s}\times\vec{t}$

I read that the magnitude of $\vec{s}\times\vec{t}$ is calculated by $rs|sin(a°-b°)|$. I'm confused as to why we can calculate it using sin(a°-b°) and what's the geometrical/physical meaning of doing it this way. Based on my knowledge, the only way I came up with is moving $\vec{t}$ so that its tail is at the head of $\vec{s}$. then we can calculate it as if calculating the torque using $\tau=Fl$ enter image description here

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  • $\begingroup$ What do you want to compute? The torque $\vec F\times \vec r$? $\endgroup$
    – kyril
    May 21, 2021 at 18:11
  • $\begingroup$ Roughly speaking, the magnitude of the cross product is the determinant of matrice given by the two vectors expressed on the plane they are defined within. And the geometrical interpretation of this determinant is that it is the area of the parallelogram given by these two vectors. Hence you have a $\sin$ function. $\endgroup$ May 21, 2021 at 18:18
  • $\begingroup$ @kyril I want to compute $\vec{s}\times\vec{t}$, but I don't understand the solution described in my question as I only know how to calculate 2 vectors as if calculating a torque... $\endgroup$
    – Alexia.
    May 21, 2021 at 18:26
  • $\begingroup$ I'm sorry I'd like to help, but I don't understand what $s$ and $t$ are... $\endgroup$
    – kyril
    May 21, 2021 at 21:15

3 Answers 3

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The magnitude of the vector product of s and t is: st[sin(angle between s and t)]. In the given problem, that angle is the difference between the two angles measured from the x axis. In your diagram, s would be the radius and t would be F, but there is no reason to assume that F is vertical, and you have not specified the angles from the x axis (assumed horizontal?).

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Well you are not wrong but it would ethically be an injustice to physics if we described it that way . This is because cross product comes first and then torque. So we say that $\tau$ is the cross product of $\vec S$ and $\vec F$ and not the other way round.$$\vec \tau = \vec S \times \vec F$$ It was simply defined that cross product of two vectors $\vec A$ and $\vec B$ is AB$sin \theta$ ($\theta$ being the angle between the two vectors) and have direction as given by screw rule because many quantities like torque were found to obey such results. Besides it is the only way where we can define pseudo vectors like torque that describe motion not along a line but along a plane and are directed perpendicular to such a plane .

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Here is another way to evaluate the vector cross product of two vectors using a determinate.

For any two vectors $\vec s$ and $\vec t$ $$\ {\vec s \times \vec t} = \begin{vmatrix} \vec i & \vec j & \vec k \\ s_x & s_y & s_z \\ t_x & t_y & t_z \end{vmatrix}$$ $\vec i$, $\vec j$, and $\vec k$ are unit vectors along the $x$, $y$, and $z$ axes of a Cartesian coordinate system. $s_x$ , $s_y$, and $s_z$ are the components of $\vec s$ along the axes, and $ t_x$ , $t_y$, and $t_z$ are the components of $\vec t$ along the axes.

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