Vector equation of a plane

To determine a plane in space we need a point and two different directions. These directions are given by two linearly independent vectors that are called director vectors of the plane.

It is important to remark that it is equivalent to have a point and two linearly independent vectors as it is to have three non aligned points. Let's see it:

If we have three points $$A, B$$ and $$C$$, we can obtain $$1$$ point and two vectors by doing: $$$\begin {array}{rcl}P&=&A \\ \overrightarrow{v}&=&\overrightarrow{AB} \\ \overrightarrow{w}&=&\overrightarrow{AC}\end{array}$$$

Obviously, if we have $$1$$ point $$P$$ and two vectors $$\overrightarrow{v}$$ and $$\overrightarrow{w}$$ can obtain three points by doing: $$$\begin{array}{rcl} A&=&P \\ B&=& P + \overrightarrow{v} \\ C&=& P+\overrightarrow{w}\end{array}$$$ Let's consider now in the reference system $$\{O; \overrightarrow{i},\overrightarrow{j},\overrightarrow{k}\}$$ the plane $$\pi$$ that goes through the point $$P$$ and has director vectors $$\overrightarrow{v}$$ and $$\overrightarrow{w}$$. We will symbolize it by $$\pi(A,\overrightarrow{v},\overrightarrow{w})$$.

As in the case of the straight line, we can express any point in the plane applying a linear combination of two governing vectors of the plane with a point in the plane.

We then know that the vector equation is: $$$P = A +\lambda \overrightarrow{v} +\mu \overrightarrow{w}$$$ which expressed in coordinates is: $$$(x,y,z)=(a_1,a_2,a_3) +\lambda \cdot (v_1,v_2,v_3)+\mu \cdot (w_1,w_2,w_3)$$$

Consider points $$A = (1,-3, 5), B = (1, 2,-1)$$ and $$C = (-2,-1, 0)$$ find the vector equation of the plane that they define.

We find the director vectors of the plane by doing: $$$\begin{array}{rcl}\overrightarrow{v}&=&\overrightarrow{AB}=B-A=(1,2,-1)-(1,-3,5)=(0,5,-6) \\ \overrightarrow{w}&=&\overrightarrow{AC}=C-A=(-2,-1,0)-(1,-3,5)=(-3,2,-5)\end{array}$$$ and like this we know that the vector equation is: $$$(x, y, z) = (1,-3, 5) + \lambda \cdot (0, 5,-6) + \mu \cdot (-3, 2,-5)$$$