What is the vector equation of a plane that contains the line `y=(3,2,1)+x(1,0,2) ` and is parallel to the z-axis?
The equation of a plane can be easily based on its relationship between two lines. In this case, we know that it contains a given line, and that it is parallel to another. We can use these facts to easily determine the coefficients of the two variables required in a plane's vector equation.
Plane equations take the following form:
`f (s,t) = vecx_1 s + vecx_2 t + vecc`
Both of the lines above can be considered "part" of the plane, because they will make 2 directions that can be combined by a constant set of coefficients (`s,t`) to produce any point on the plane. This fact makes it very easy to make a vector equation for the plane given the two lines! We'll start by incorporating the vector form of the line given as part of the plane:``
`f(s,t) = (1,0,2)s + vecx_2t+vecc`
Next, because we know the plane is parallel to the z-axis, we can incorporate a vector equation for the z-axis:
`vecz(x) = (0,0,1)x`
Now, we just substitute the vector we used to define the z-axis into our plane equation:
`f (s,t)=(1,0,2)s+(0,0,1)t + vecc`
Finally, to make sure the line, itself, is included in the plane, we must add in the constant vector offset that is already given in the line:
`f(s,t) = (1,0,2)s + (0,0,1)t + (3,2,1)`
This last step completes the equation of the plane. We can see that the given line is part of the plane by setting `t=0` and getting the line's equation. We can also see that the plane is parallel to the z-axis by noting that the two can never intersect (note that all `y` values must be 2 on the plane, but 0 on the z-axis).
Therefore, our plane equation has been found:
`f(s,t) = (1,0,2)s+(0,0,1)t + (3,2,1)`