9 Solving Two Vectors in the 𝑥, 𝑦, and 𝑧 Axis

To solve two vectors the same rules from part one applies to these calculations. We will be given two magnitudes, both with two angles each. The first step is to find coordinates for said vectors, then combine the like terms and calculate the resultant magnitude. Finally, we calculate the resultant angles. Furthermore, remember to keep negative values negative, and positive values positive.

Our given values will be:

[latex]\vec{f_1}=2[/latex] units,

[latex]\theta_{1xy}={18}^\circ[/latex],

[latex]\theta_{1xyz}={37}^\circ[/latex],

And,

[latex]\vec{f_2}=2[/latex] units,

[latex]\theta_{2xy}={198}^\circ[/latex],

[latex]\theta_{2xyz}={217}^\circ[/latex].

Our first step is to combine the like terms, as we cannot simply add the resultants and angles together to secure our solution. We will be using methods previously discussed to solve for our three values each:

[latex]f_{1xy}=1.60[/latex],

[latex]f_{1x}=1.52[/latex],

[latex]f_{1y}=0.50[/latex],

[latex]f_{1z}=1.20[/latex],

And,

[latex]f_{2xy}=-1.60[/latex],

[latex]f_{2x}=-1.52[/latex],

[latex]f_{2y}=-0.50[/latex],

[latex]f_{2z}=-1.20[/latex].

I am sure that you have by now noticed that these numbers completely cancel out to zero, to which you would be correct. Due to the visually complex nature of the upcoming equations, I find it necessary to present a problem that can be solved mentally beforehand so as to prove to you, the reader, that these equations do indeed function as intended within my stated hyperplane. Seeing how they work before indulging into the actual equations will be an immense help during the following chapter.

After combining all the values, we are left with:

[latex]R_x=0[/latex],

[latex]R_y=0[/latex],

[latex]R_z=0[/latex].

Due to this vector not existing at all, there should be absolutely no direction as there is nothing to have direction. We can verify this by using the tangent law to solve for the angle:

[latex]\theta_{Rxy}=0[/latex],

[latex]\theta_{\vec{R}xyz}=0[/latex].

Now that we have a surface level understanding of how the system and its equations work, let us run another one that will give us an actual value for the resultant and its angles.

Our given values will be:

[latex]\vec{f_1}=18[/latex] units,

[latex]\theta_{1xy}={17}^\circ[/latex],

[latex]\theta_{1xyz}={48}^\circ[/latex],

And,

[latex]\vec{f_2}=19[/latex] units,

[latex]\theta_{2xy}={127}^\circ[/latex],

[latex]\theta_{2xyz}={343}^\circ[/latex].

As per the last calculations we must solve for the like terms as a means to combine them and ascertain the resultant and its angles. Using the sine and cosine trigonometric functions we come to values of:

[latex]f_{1xy}=12.05[/latex] units,

[latex]f_{1z}=13.38[/latex] units,

And,

[latex]f_{2xy}=18.17[/latex] units,

[latex]f_{2z}=-5.56[/latex] units.

From these values we can calculate our [latex]x[/latex] and [latex]y[/latex] values using the same trig functions,

[latex]f_{1x}=11.52[/latex] units,

[latex]f_{1y}=3.50[/latex] units,

And,

[latex]f_{2x}=-10.93[/latex] units,

[latex]f_{2y}=14.51[/latex] units.

We can now combine our values to get the resultant coordinates,

[latex]R_x=0.59[/latex] units,

[latex]R_y=18.01[/latex] units,

[latex]R_z=7.82[/latex] units.

From here we can plug the [latex]x[/latex] and [latex]y[/latex] values into pythagorean theorem to get our [latex]R_{xy}[/latex],

[latex]R_{xy}=18.02[/latex]

And using this we can utilize the same method to get our resultant magnitude, except this time we are using our [latex]z[/latex] value:

[latex]{\vec{R}}_{xyz}=19.64[/latex]

From here we now need to calculate our two angles, this can be done by using the tangent function for reasons previously stated^[1], we will begin with the value in the [latex]xy[/latex] plane:

[latex]\left(tan\theta_{Rxy}=\frac{18.01}{0.59}\right)\Rightarrow\left(\theta_{Rxy}={tan}^{-1}\left(\frac{18.01}{0.59}\right)\right)[/latex]

Which ends up being,

[latex]\theta_{Rxy}={88.12}^\circ[/latex]

Now we can solve for the resultant angle in the hyperplane. We will use the same method except that this time we will be using our [latex]R_{xy}[/latex] and our [latex]R_z[/latex].

[latex]\left(tan\theta_{\vec{R}xyz}=\frac{7.82}{18.02}\right)\Rightarrow\left(\theta_{\vec{R}xyz}={tan}^{-1}\left(\frac{7.82}{18.02}\right)\right)[/latex]

Solves as [latex]\theta_{\vec{R}xyz}={23.46}^\circ[/latex].