10 The Basic Formulas for Three-Dimensional Vectors

Alright, hold on. What just happened there? What was the process we took, and why use that one? The process we took is the simplest approach to my system and can be used to perform my resultant equation on paper. The equation we will begin by talking about will be the simplest form of the vector resultant equation.

[latex]{\vec{R}}_{xyz}=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2+\left(R_z\right)^2}[/latex]

That is it. It is just the first formula with an extra value, due to there being an extra axis. While making this formula, I began by thinking about the process I take as a means to solve for three-dimensional linear vector addition. I reminded myself of the solving of axis and shadows. I then input them into an equation that reads as follows:

[latex]{\vec{R}}_{xyz}=\sqrt{\left(R_{xy}\right)^2+\left(R_z\right)^2}[/latex]

And can be expressed as:

[latex]R_{xy}=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}[/latex]

But of course, I would not settle on two equations for a single value, as that would be counterintuitive towards the entirety of this paper. So, I combined them. And by combining them I ended up with:

[latex]{\vec{R}}_{xyz}=\sqrt{\left(\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}\right)^2+\left(R_z\right)^2}[/latex]

Which makes complete sense. This formula, though correct, can be simplified even further. Please ignore the fact that this took me over three weeks to realize this. Can you see what I could not? The square and the square root within the initial square cancel each other out, meaning that the first equation I presented in this chapter is the correct one.

But let us dive deeper. Just the same as the last one the axis of the resultants are equal to the sum of the same axis. This can be written as,

[latex]f_{1x}+f_{2x}=R_x[/latex]

[latex]f_{1y}+f_{2y}=R_y[/latex]

[latex]f_{1z}+f_{2z}=R_z[/latex]

Each of these “[latex]f[/latex]” values can be calculated using sine and cosine. We can use our initial magnitude to solve for the force in the [latex]xy[/latex] plane, and the force in the [latex]z[/latex] plane. Then utilizing our newly calculated force in the [latex]xy[/latex] plane, solve for the force in the [latex]x[/latex] and [latex]y[/latex]. These can be expressed as,

[latex]\left(cos\theta_{1xyz}=\frac{f_{1xy}}{\vec{f_1}}\right)\Rightarrow f_{1xy}[/latex]

[latex]\left(cos\theta_{1xy}=\frac{f_{1x}}{f_{1xy}}\right)\Rightarrow f_{1x}[/latex]

[latex]\left(sin\theta_{1xy}=\frac{f_{1y}}{f_{1xy}}\right)\Rightarrow f_{1y}[/latex]

[latex]\left(sin\theta_{1xyz}=\frac{f_{1z}}{\vec{f_1}}\right)\nRightarrow f_{1z}=\sqrt{\left(\vec{f_1}\right)^2+\left(f_{1xy}\right)^2}[/latex]

And the same for other numbers of vectors.

You will notice that the final equation, the labeled Pythagorean theorem, is incorrect. As per the description of Pythagorean theorem in chapter four, the equation should not be written this way. Our [latex]\vec{f_1}[/latex] is the hypotenuse, and the equation states that the area of side [latex]a[/latex] plus the area of side [latex]b[/latex] is equal to the area of hypotenuse [latex]c[/latex]. Therefore, the correct way to express these equations is,

[latex]f_z=\sqrt{\left(\vec{f}\right)^2-\left(f_{xy}\right)^2}[/latex]

As this states that side [latex]b[/latex] is equal to the root of the sum of the square of hypotenuse [latex]c[/latex] minus the square of side [latex]a[/latex].

For the angles, it is pretty similar to the previous part. We use tangent to calculate our angles, keeping in mind to use the applicable axis accordingly. The basic way of writing these equations is,

[latex]\left(tan\theta_{Rxy}=\frac{R_y}{R_x}\right)\Rightarrow\left(\theta_{Rxy}={tan}^{-1}\left(\frac{R_y}{R_x}\right)\right)[/latex]

[latex]\left(tan\theta_{\vec{R}xyz}=\frac{R_z}{R_{xy}}\right)\Rightarrow\left(\theta_{\vec{R}xyz}={tan}^{-1}\left(\frac{R_z}{R_{xy}}\right)\right)[/latex]

This is directly the exact same method of finding the angles as in part one, the exception being the latter equation uses the “shadow” of the vector under noon sun.