Equations of a plane

An equation, like a line, can be expressed in several ways. We start with the simplest, parametric expression of a plane.

The parametric expression of a plane

Let's first recall how we determined the parametric equation of the line p. We chose a point A that crossed the line p and then a direction vector $\vec{\mathbf{u}}$. If we added some t-fold of the direction vector to the point A, we got some point on the line p. For every point X that lies on the line, we got the equation

$$ X = A + t\cdot \vec{\mathbf{u}},\quad t\in \mathbb{R}. $$

By similar reasoning, we can arrive at the equation of the plane. To describe the plane, we need two different nonzero vectors that are noncollinear, i.e., that are not parallel, and of course the point A, through which the plane passes. The vectors must not be collinear because they do not lie on the same line. One line does not define a plane, one line can be contained in infinitely many planes.

We can go about it another way - we are able to define a line using two distinct points A, B. However, these points are crossed by infinitely many planes - but if we add one distinct point to these points C, then we have uniquely defined a plane.

Thus, if we have three distinct points A, B, C, we can construct two distinct vectors $\vec{\mathbf{u}}=\vec{AB}$ and $\vec{\mathbf{v}}=\vec{AC}$. These two vectors and, for example, the point A define a plane. Recall the addition of vectors:

We see that by summing two vectors we can get a vector that has coordinates "between" the two vectors. By successive addition of different multiples of the vectors $\vec{\mathbf{u}}$ and $\vec{\mathbf{v}}$ we are then able to fill the whole plane. We can thus write a parametric expression of the plane that is determined by the vectors $\vec{\mathbf{u}}, \vec{\mathbf{v}}$ and that passes through the point A:

$$ X = A + t\cdot \vec{\mathbf{u}} + s\cdot \vec{\mathbf{v}}, \quad t, s \in \mathbb{R} $$

where X is some point on the plane. For each point of the plane, we are able to find such t and s that the equation holds. We can still break this equation down into a system of equations. We assume that X[x, y, z], A[a₁, a₂, a₃], $\vec{\mathbf{u}}=(u_1, u_2, u_3)$, and $\vec{\mathbf{v}}=(v_1, v_2, v_3)$.

\begin{eqnarray} x &=& a_1 + t \cdot u_1 + s \cdot v_1 \\ y &=& a_2 + t \cdot u_2 + s \cdot v_2 \\ z &=& a_3 + t \cdot u_3 + s \cdot v_3 \\ \end{eqnarray}

Example

Determine whether the point Q[0,3,5] or the point A[2,3,4], B[3,7,−1], C[−5,4,4] lies in the plane defined by .

The first thing to do is to construct the equation of the plane. Determine the vectors $\vec{\mathbf{u}}$ and $\vec{\mathbf{v}}$, with $\vec{\mathbf{u}}=\vec{AB}$ and $\vec{\mathbf{v}}=\vec{AC}$. This will be true:

\begin{eqnarray} \vec{\mathbf{u}}&=&B-A=[3,7,-1]-[2,3,4]=[1,4-5]\\ \vec{\mathbf{v}}&=&C-A=[-5,4,4]-[2,3,4]=[-7,1,0] \end{eqnarray}

The equation passes through the point A, so we write a system of equations:

\begin{eqnarray} x &=& 2 + t \cdot 1 - s \cdot 7 \\ y &=& 3 + t \cdot 4 + s \cdot 1 \\ z &=& 4 - t \cdot 5 + s \cdot 0 \\ \end{eqnarray}

Now we find out if the point Q[0,3,5] passes through this plane. We do this by just plugging the numbers 0,3,5 after x,y,z and solving the system of equations. So we get the system:

\begin{eqnarray} 0 &=& 2 + t \cdot 1 - s \cdot 7 \\\ 3 &=& 3 + t \cdot 4 + s \cdot 1 \\ 5 &=& 4 - t \cdot 5 + s \cdot 0 \\ \end{eqnarray}

From the last equation we can isolate t:

\begin{eqnarray} 5 &=& 4 - t \cdot 5 + s \cdot 0 \\\ 5 &=& 4 - t \cdot 5\\ 1 &=& - t \cdot 5\ t \cdot 5 &=& -1\ t &=& -\frac{1}{5} \end{eqnarray}

Plug t into the second equation:

\begin{eqnarray} 3 &=& 3 + t \cdot 4 + s \cdot 1 \\\ 3 &=& 3 - \frac{1}{5} \cdot 4 + s \cdot 1 \\ 0 &=& -\frac{4}{5} + s \cdot 1 \\ s &=& \frac{4}{5} \end{eqnarray}

Finally, we plug the values of t and s into the first equation to see if the equation makes sense:

\begin{eqnarray} 0 &=& 2 + t \cdot 1 - s \cdot 7 \cdot 7 \cdot 0 &=& 2 - \frac{1}{5} \cdot 1 - \frac{4}{5} \cdot 7 \cdot 7 \cdot 0 &=& 2 - \frac{1}{5}=& \frac{10}{5} - \frac{1}{5} - \frac{28}{5}=0 &=& -\frac{19}{5} \end{eqnarray}

The equation doesn't make sense, the whole system has no solution there, the point Q[0,3,5] is not a point of the plane.