12 Solving Many Vectors in the 𝑥, 𝑦, and 𝑧 Axis

Identically to last time we will be running a question with multiple variables, and we will use the summed variations of my formulas to solve them. For this question we will be solving for three separate vectors, and their resultant in the [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex] axis, our givens:

[latex]\vec{f_1}=42[/latex]

[latex]\vec{f_2}=13[/latex]

[latex]\vec{f_3}=37[/latex]

[latex]\theta_{1xy}=75°[/latex]

[latex]\theta_{2xy}=265°[/latex]

[latex]\theta_{3xy}=163°[/latex]

[latex]\theta_{1xyz}=342°[/latex]

[latex]\theta_{2xyz}=27°[/latex]

[latex]\theta_{3xyz}=222°[/latex]

The units are unimportant for this question as they are all the same. The equations that we will use to solve for the addition of the vector magnitudes can now be selected. We will be using my summed formulas of course, as we have more than two vectors to work with.

[latex]{\vec{R}}_{xyz}=\sum_{i=1}^{n}\sqrt{\left(\vec{f_i}sin\theta_{ixyz}\right)^2+\left(\left(\vec{f_i}cos\theta_{ixyz}\right)cos\theta_{ixy}\right)^2+\left(\left(\vec{f_i}cos\theta_{ixyz}\right)sin\theta_{ixy}\right)^2}[/latex]

This is the base equation we will be using. Just like last time we will be plugging in our values into similar spots on the summation relative to their number. Due to the size of this equation, I will not be writing down all three vector values^[1], instead I will skip to the part where we combine the magnitudes. All values will be rounded to two decimal places.

[latex]\vec{R_{xyz}}=\sqrt{\left(-12.98+5.9-24.75\right)^2+\left(10.34-1.01+26.29\right)^2+\left(38.58-11.54-8.04\right)^2}[/latex]

Synonymous with chapter six, we combine what is inside the brackets. This will give us our [latex]z[/latex], [latex]x[/latex], and [latex]y[/latex] values, respectively.

[latex]\vec{R_{xyz}}=\sqrt{\left(-31.83\right)^2+\left(35.62\right)^2+\left(19\right)^2}[/latex]

Based on these values we can determine the quadrant the vector resultant will lie in. The resultant will be in quadrant four.

Now we can solve for the resultant angles. We will start off simply be solving for the resultant angle in the [latex]xy[/latex] plane. Based on the summed formula in the previous chapter, the angle is,

[latex]\theta_{\vec{R}xy}=\tan^{-1}{\left(\frac{19}{35.62}\right)}[/latex]

[latex]\theta_{\vec{R}xy}=28.07°[/latex]

We can put this in standard form relative to quadrant four: based on the “swivel” angle of the resultant, this is equal to [latex]0°[/latex] (starting angle due to being in the first quadrant) [latex]+\ 28.07°=28.07°[/latex].

Now using the same calculated values, solve for our resultant angle in the [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex] axis. The secondary summed formula is to be employed to do this but, just like last time we have already solved for our vector coordinates. It is as easy as plugging our values into the base equation to solve for our angle.

[latex]\theta_{\vec{R}xyz}=\tan^{-1}{\left(\frac{-31.83}{\sqrt{\left(35.62\right)^2+\left(19\right)^2}}\right)}[/latex]

[latex]\theta_{\vec{R}xyz}=-30.23°[/latex]

Finally, this needs to be put into standard form. Since this angle is in the fourth quadrant, subtraction from a full rotation is to be utilized to gain the values, [latex]360°-30.23°=329.77°[/latex].

Vectors are typically solved with tables to spot mistakes along the way. I argue however that if you correctly plug your values into a calculator, it will not make a mistake, thus providing a correct resultant. Not only does this save time, but it reduces the number of steps needed, reducing the likelihood of human error. Calvin’s two- and three-dimensional linear vector addition formulas provide a more efficient and elegant method of solving for the redirection of vectors. The contents of the text prove that the presented formulas work effectively and provide a clear route for anyone to follow.