Two vector equations of a plane, and how to obtain Cartesian form

A point and a normal vector

Let a plane pass through a point with position vector $\mathbf a$, and let $\mathbf n\ne\mathbf0$ be a normal vector. A point with position vector $\mathbf x$ lies on the plane exactly when

$$\boxed{\mathbf n\cdot(\mathbf x-\mathbf a)=0.}$$

Every displacement within the plane is perpendicular to its normal. Conversely, the points satisfying this equation form the plane through $\mathbf a$ perpendicular to $\mathbf n$. Requiring a nonzero normal prevents the equation from degenerating into $0=0$ for all of space.

A point and two independent directions

Alternatively, let $\mathbf b,\mathbf c$ be linearly independent vectors parallel to the plane. Then its points are exactly

$$\boxed{\mathbf x=\mathbf a+s\mathbf b+t\mathbf c,\qquad s,t\in\mathbb R.}$$

The two direction vectors span all displacements along the plane, and adding $\mathbf a$ places it through the chosen point. Independence matters: parallel directions would span only a line, and zero directions can degenerate further.

Example 1: expand a normal equation

Problem. Find the equation of the plane through $A=(1,2,3)$ with normal vector $(1,2,-1)$.

Let $(x,y,z)$ be a general point on it. The normal equation is

$$(1,2,-1)\cdot(x-1,y-2,z-3)=0.$$

Expand the dot product:

$$x-1+2(y-2)-(z-3)=0.$$

Therefore

$$\boxed{x+2y-z=2.}$$

Substituting $A$ checks the constant: $1+2\cdot2-3=2$. The coefficient vector of the Cartesian equation is the given normal.

Example 2: convert two spanning directions

Problem. Find the equation of the plane through $A=(1,2,3)$ parallel to $(2,5,1)$ and $(3,2,1)$.

The parametric form is

$$(x,y,z)=(1,2,3)+s(2,5,1)+t(3,2,1).$$

You can eliminate $s,t$ from the three coordinate equations. Another route is to find a vector perpendicular to both directions.

Find and verify a normal

The cross product gives

$$(2,5,1)\times(3,2,1)=(3,1,-11).$$

Whether obtained by a cross product or by solving two equations, verify the result directly:

$$(3,1,-11)\cdot(2,5,1)=6+5-11=0,$$ $$(3,1,-11)\cdot(3,2,1)=9+2-11=0.$$

This nonzero vector is therefore normal to the plane. Using the known point gives

$$3(x-1)+(y-2)-11(z-3)=0,$$

hence

$$\boxed{3x+y-11z=-28.}$$

The same result by eliminating parameters

From $z=3+s+t$, we have $t=z-3-s$. Substitute into $x=1+2s+3t$:

$$s=3z-x-8.$$

Then $y=2+5s+2t=3s+2z-4=11z-3x-28$, which rearranges to the same equation. Thus the two descriptions are equivalent, not competing definitions.

How to present a solution clearly

If a vector method lies outside the syllabus assumed by an examination, the two dot-product checks provide a short elementary justification of the proposed normal. There is no need to conceal how it was found. Always state a point on the plane and a nonzero normal, or a point and two independent directions.

The projection and reflection formulas in orthogonal projection vectors use the same normal-vector description.