1.6 - Equations of planes

1.6 - Equations of planes

A plane can be described by a perpendicular direction or by two directions along its surface. You will build Cartesian, scalar-product and parametric equations, then change between them while checking that they describe the same points.

A plane from its normal

How can one equation describe every point on a flat surface in three dimensions? Start with the horizontal plane z=3z=3. Every joining vector between its points has zero vertical component, so it is perpendicular to k\mathbf k. The same idea describes a tilted plane.

Let AA be a fixed point on the plane, with position vector a\mathbf a. A point with position vector r\mathbf r belongs to the plane precisely when its displacement r−a\mathbf r-\mathbf a is perpendicular to a fixed nonzero normal vector n\mathbf n.

Normal equation of a plane

(r−a)⋅n=0⟺r⋅n=p,p=a⋅n.(\mathbf r-\mathbf a)\cdot\mathbf n=0 \quad\Longleftrightarrow\quad \mathbf r\cdot\mathbf n=p, \qquad p=\mathbf a\cdot\mathbf n.

If n=(a,b,c)≠0\mathbf n=(a,b,c)\ne\mathbf0, this becomes the Cartesian equation

ax+by+cz=p.ax+by+cz=p.

The normal need not be a unit vector. Multiplying the entire equation by a nonzero constant gives the same plane.

Not in the formula booklet: learn it.

Here pp is a scalar, not a point. The vector a\mathbf a locates the plane; n\mathbf n fixes its orientation. A zero normal gives either a tautology or an impossibility, rather than an equation of a plane.

Worked example 1: a point and a normal

Question. Find an equation of the plane through A=(1,2,3)A=(1,2,3) with normal (2,−1,2)(2,-1,2).

Plan. Find the normal's scalar product with AA, then use the same value for every point of the plane.

p=2(1)−2+2(3)=6r⋅(2,−1,2)=6.\begin{aligned} p&=2(1)-2+2(3)=6\\ \mathbf r\cdot(2,-1,2)&=6. \end{aligned}

Answer. 2x−y+2z=62x-y+2z=6.

Check. Substitution of AA gives 6=66=6. Also, (1,2,0)(1,2,0) and (0,2,1)(0,2,1) have zero scalar product with the normal; adding either to AA preserves the equation, as movement along a plane should.

Now change only the length and sense of the normal. The surface itself must stay fixed.

Worked example 2: equivalent normal equations

Question. Find the plane through (1,2,3)(1,2,3) with normal (−4,2,−4)(-4,2,-4), and compare it with Worked example 1.

Plan. Calculate the new constant, then simplify the whole equation.

p=−4(1)+2(2)−4(3)=−12−4x+2y−4z=−122x−y+2z=6divide every term by −2.\begin{aligned} p&=-4(1)+2(2)-4(3)=-12\\ -4x+2y-4z&=-12\\ 2x-y+2z&=6 &&\text{divide every term by }-2. \end{aligned}

Answer. It is the same plane, 2x−y+2z=62x-y+2z=6.

Check. The new normal is −2-2 times the old one, and the new constant is also −2-2 times the old constant. The equations therefore have exactly the same solution set.

Scaling just the left-hand side would usually move the plane. The constant must scale too.

Watch out

A normal vector alone describes a family of parallel planes. A point, or the constant in the equation, is also needed to select one plane.

Use both pieces of information in this exercise.

Two directions in a plane

You can also build a plane by moving from a fixed point in two independent directions. One parameter would describe only a line; two parameters allow movement across a surface.

For example, start at (1,0,2)(1,0,2), move λ\lambda copies of (1,1,0)(1,1,0), then μ\mu copies of (0,1,1)(0,1,1). Each pair of real parameters gives a point of one plane.

Parametric equation of a plane

r=a+λb+μc,λ,μ∈R.\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c, \qquad \lambda,\mu\in\mathbb R.

The direction vectors b,c\mathbf b,\mathbf c must be nonzero and not parallel. Equivalently, b×c≠0\mathbf b\times\mathbf c\ne\mathbf0.

Not in the formula booklet: learn it.

The parameters vary independently. If c\mathbf c were a multiple of b\mathbf b, both movements would be along one line. Two symbols alone do not guarantee a plane.

Worked example 3: using two given directions

Question. A plane passes through (1,0,2)(1,0,2) and is parallel to (1,1,0)(1,1,0) and (0,1,1)(0,1,1). Give a parametric equation and determine whether Q=(3,1,1)Q=(3,1,1) lies in it.

Plan. Use the point and directions, then find one pair of parameters satisfying all three coordinates of QQ.

r=(102)+λ(110)+μ(011).\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix} +\lambda\begin{pmatrix}1\\1\\0\end{pmatrix} +\mu\begin{pmatrix}0\\1\\1\end{pmatrix}.

The first coordinate of QQ gives 1+λ=31+\lambda=3, so λ=2\lambda=2. The third gives 2+μ=12+\mu=1, so μ=−1\mu=-1. The second then gives λ+μ=1\lambda+\mu=1, as required.

Answer. The displayed equation describes the plane, and QQ lies on it with (λ,μ)=(2,−1)(\lambda,\mu)=(2,-1).

Check. Direct vector addition gives (1,0,2)+2(1,1,0)−(0,1,1)=(3,1,1)(1,0,2)+2(1,1,0)-(0,1,1)=(3,1,1). The direction vectors are not multiples, so they really span a plane.

Keep the same surface, but now obtain the directions from three points instead of receiving them directly.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account