Parametric Equation of a Plane

There are two forms of equation of a plane : a vector form and a Cartesian form. 

Further, there are two kinds of vector equation forms : a vector equation deduced using any normal(orthogonal or perpendicular) vector to the plane, and the other is called the parametric equation form which results from combination of vectors parallel to the plane. We have already looked at the former in one of my other posts. We will look at the latter now.

The following analogy of a clock might help to get a basic idea in the beginning. In one complete revolution from 12 to 12, a minute hand(or any other hand of clock) points in every direction in the plane of the clock : N, NE, E, SE, etc… Or we can say that it points in every direction in the plane of the wall to which the clock is hung.

To find a parametric equation of a plane in 3D space, we need a vector(similar to the minute-hand arrow), call it a variable vector, which can point in every possible direction in the given plane but must always stay parallel to the plane.

Furthermore, we also want this vector to take all possible magnitudes, i.e, to shorten or lengthen to any lengths as desired(no restrictions). Its magnitude and direction are dictated by certain “parameters”(two in number) which are just numbers. By assigning values to the parameters we can adjust the magnitude of the vector arrow to any value and make it point in any direction around the plane(while always remaining parallel to the plane). 

To define such a variable vector we first need two non-parallel vectors but both parallel to the plane. The vectors should not be parallel to each other but both should be parallel to the plane. Suppose there is a plane P and two non-parallel vectors are $\overrightarrow{a}$ and $\overrightarrow{b}$. Their resultant vector is $\overrightarrow{b}+\overrightarrow{a}$ as shown below, which will also be parallel to the plane, and which will have a certain magnitude and a certain direction in the plane. Let this be called $\overrightarrow{c}$.

Right now the coefficients of both $\overrightarrow{a}$ and $\overrightarrow{b}$ is 1. If we multiply $\overrightarrow{a}$ and $\overrightarrow{b}$ by any other positive numbers, say $\alpha$ and $\beta$, their magnitudes will change but their directions will not(because the coefficients are positive). If either of $\alpha$ or $\beta$ is negative then the direction of $\alpha \overrightarrow{a}$ or $\beta \overrightarrow{b}$ will be reversed. Any change in their coefficients will cause their resultant $\overrightarrow{c}$ to also change in magnitude and/or direction. 

$\alpha \overrightarrow{a}$ and $\beta \overrightarrow{b}$, each can only go two ways, but the resultant $\overrightarrow{c}$ can go to any direction(but always remaining parallel to the plane). Moreover, the resultant vector $\overrightarrow{c}$ can assume any size, as big or as small as we want. In other words, there will always exist $\alpha$ and $\beta$ values by which the resulting vector $\overrightarrow{c}$ can be adjusted to the magnitude and direction of any other vector parallel to the plane.

The resultant vector $\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}$ is our “variable vector” that we are after, and $\alpha$ and $\beta$ are the parameters which set the magnitude and direction of this vector. Also, $\alpha$ and $\beta$ can be positive, negative or zero. 

Let A(p, q, r) be some fixed point in the plane, and let B(x, y, z) be an arbitrary point(it can be anywhere in the plane). We will define yet another vector, $\overrightarrow{AB}$.

$\overrightarrow{AB}=(x-p)\hat{i}+(y-q)\hat{j}+(z-r)\hat{k}$

Clearly, this is also a vector that is parallel to the plane, and will always be so regardless of where the arbitrary point B is in the plane. But we just saw the magnitude and direction of $\overrightarrow{c}$ can be adjusted to the magnitude and direction of any vector parallel to the plane through alpha and beta. This means for every point (x, y, z) in the plane there will exist $\alpha$ and $\beta$ such that $\overrightarrow{c}$ and \(\overrightarrow{AB}\) are equal. 

And because $\overrightarrow{c}$ itself always stays parallel to the plane for all $\alpha$ and $\beta$, the converse of the above is true too. That is, for any values of $\alpha$ and $\beta$ there will be a point (x, y, z) in the plane such that the vectors are equal. There is a 1-1 relation between the three coordinates and the two parameters. So we can set both the vectors to be equal.

$(x-p)\hat{i}+(y-q)\hat{j}+(z-r)\hat{k}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}$

The above is referred to as the parametric equation of the plane. 

So to sum up, to find parametric form of equation, we need to have :

Coordinates of any one point(fixed) in the plane, and

Two non-parallel vectors both parallel to the plane

As an example, consider a plane passing through (1,0,0), (0,1,0) and (0,0,1). And we want to determine its parametric equation. 

Firstly, we can take any of the three given points as our fixed point required to find the equation. Let’s go with B(0,1,0). Let A(x, y, z) be an arbitrary point in the plane.

$⟹\overrightarrow{AB}=(x-0)\hat{i}+(y-1)\hat{j}+(z-0)\hat{k}$

For $\overrightarrow{c}$, we need any two vectors non-parallel to each other but both parallel to the plane. They can be easily found from the three given points in the plane. Define a vector between any two given points, and another vector between another pair of points. 

$⟹\overrightarrow{a}=(1-0)\hat{i}+(0-1)\hat{j}+(0-0)\hat{k}$

                          $=1\hat{i}+(-1)\hat{j}+0\hat{k}$

(We have defined this vector from (0,1,0) to (1,0,0). But we could have gone the other way. It doesn’t matter what vector we choose as long as it is parallel to the plane)

Also,

$\overrightarrow{b}=(0-0)\hat{i}+(0-1)\hat{j}+(1-0)\hat{k}$

                          $=0\hat{i}+(-1)\hat{j}+1\hat{k}$

Note that here we can say that vectors a and b are not parallel to each other because it is obvious that the three points are non-collinear(cannot be all on a single straight line). But we can also check this using the knowledge that parallel vectors are scalar multiples of each other. That is, if they were parallel, then there would exist a scalar ‘s’ such that $\overrightarrow{a}=s\overrightarrow{b}$. 

<1, -1, 0> = s <0, -1, 1> 

Clearly there is no number ‘s’ which when multiplied with 0 gives 1, which concludes that the vectors are not parallel to each other. 

Therefore, the parametric equation of the plane is :

$(x-0)\hat{i}+(y-1)\hat{j}+(z-0)\hat{k}=\alpha (1\hat{i}+(-1)\hat{j}+0\hat{k})+\beta (0\hat{i}+(-1)\hat{j}+1\hat{k})$