If $a,b,c$ are three non coplanar vectors, then prove that the vector equation $r=(1-p-q)\vec a+p\vec b +q\vec c$ represents a plane

Question

Let $r=x\vec a + y\vec b+z\vec c$

Comparing with the given equation, we obtain $$x+y+z=1$$ which is a plane

What I don’t understand is how does this say $r$ is a plane, since r is actually $x \vec a+y\vec b+z\vec c$

Answer 1

Rearrange to $$\vec r = \vec a + p(\vec b-\vec a) +q (\vec c-\vec a)$$ And this is of the form of the plane equation $$\vec r= \vec p + \lambda \vec s +\mu \vec q $$ More precisely, this is the plane that passes through the point with position vector $\vec a$ and with normal vector $(\vec b-\vec a)\times (\vec c-\vec a)$.

Answer 2

A plane $\vec r=(x,y,z)$ that passes the point $\vec a$ and with normal vector $\vec n$ is $$(\vec r -\vec a)\cdot \vec n =0\tag1$$

Observe that the given vector can be rearranged as

$$\vec r -\vec a = p(\vec b-\vec a) +q (\vec c-\vec a)$$ which satisfies the plane equation (1) with the choice of $\vec n =(\vec b-\vec a)\times (\vec c-\vec a)$.