Planes Through Three Points

From Three Points to a Plane

A plane can be specified by a point on it and a normal vector. When three points are given instead, the key is to turn their positions into two directions lying in the plane. The cross product of those directions gives a normal vector, which can then be used to write the plane equation.

The method depends on the points not being collinear. If they are collinear, the two direction vectors point along the same line and their cross product is zero. There is then no unique plane determined by the points.

The Construction

Definition: Three points in space are noncollinear if they do not all lie on one line. Given noncollinear points \(P\), \(Q\), and \(R\), form the direction vectors \(\overrightarrow{PQ}=Q-P\) and \(\overrightarrow{PR}=R-P\). Their cross product is nonzero and perpendicular to both directions, so it is a normal vector to the plane through the points.

Write the coordinates as \(P=(x_1,y_1,z_1)\), \(Q=(x_2,y_2,z_2)\), and \(R=(x_3,y_3,z_3)\). Subtract coordinates to form the two vectors. The order of the subtraction matters for the direction of each vector, but reversing a vector only changes the sign of the resulting normal; either sign gives an equation for the same plane.

Once a normal vector \(\mathbf{n}=(A,B,C)\) is known, the plane through \(P\) has point-normal equation \(\mathbf{n}\cdot((x,y,z)-P)=0\). In coordinates, this is \(A(x-x_1)+B(y-y_1)+C(z-z_1)=0\). Expanding gives the general form \(Ax+By+Cz+D=0\), where \(D\) is determined by the point.

Choose a common starting point.
For example, use \(P\) to form \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\).

Take the cross product.
Compute \(\mathbf{n}=\overrightarrow{PQ}\times\overrightarrow{PR}\). If \(\mathbf{n}=\mathbf{0}\), the points are collinear and do not determine a unique plane.

Write the point-normal equation.
Use \(\mathbf{n}\) as the normal and any one of the three points as the point on the plane.

Expand and verify.
Write the equation in general form if desired, then substitute each given point to check that it satisfies the equation.

Why the Cross Product Works

Let \(\mathbf{u}=\overrightarrow{PQ}=(u_1,u_2,u_3)\) and \(\mathbf{v}=\overrightarrow{PR}=(v_1,v_2,v_3)\). Their cross product is

\mathbf{n}=\mathbf{u}\times\mathbf{v} =(u_2v_3-u_3v_2,\;u_3v_1-u_1v_3,\;u_1v_2-u_2v_1).

Its perpendicularity to \(\mathbf{u}\) follows by taking the dot product and expanding. Every term cancels with an equal term of opposite sign:

\begin{aligned} \mathbf{n}\cdot\mathbf{u} &=u_1(u_2v_3-u_3v_2)+u_2(u_3v_1-u_1v_3)+u_3(u_1v_2-u_2v_1)\\ &=u_1u_2v_3-u_1u_3v_2+u_2u_3v_1-u_1u_2v_3 +u_1u_3v_2-u_2u_3v_1\\ &=0. \end{aligned}

The same cancellation gives \(\mathbf{n}\cdot\mathbf{v}=0\): its expanded terms are \(u_2v_3v_1-u_3v_2v_1+u_3v_1v_2-u_1v_3v_2+u_1v_2v_3-u_2v_1v_3\), which cancel in pairs. Thus \(\mathbf{n}\) is perpendicular to both directions in the plane.

For noncollinear points, \(\mathbf{u}\) and \(\mathbf{v}\) are not parallel, so their cross product is nonzero. The plane through \(P\) in those two independent directions has normal \(\mathbf{n}\). A point \(X=(x,y,z)\) lies in that plane exactly when its displacement from \(P\) is perpendicular to the normal, giving \(\mathbf{n}\cdot(X-P)=0\). This establishes the point-normal equation used in the construction.

Theorem: Three noncollinear points determine a plane. Form two direction vectors from one of the points, take their cross product to obtain a nonzero normal vector, and substitute that normal and any given point into the point-normal equation. The resulting plane equation is satisfied by all three points.

Worked Constructions

Worked Example: A Plane in General Form

Find the plane through \(P=(1,0,2)\), \(Q=(3,1,1)\), and \(R=(0,2,4)\).

Form two vectors from \(P\):

\overrightarrow{PQ}=(3-1,\;1-0,\;1-2)=(2,1,-1), \qquad \overrightarrow{PR}=(0-1,\;2-0,\;4-2)=(-1,2,2).

Take their cross product, showing each component:

\begin{aligned} \overrightarrow{PQ}\times\overrightarrow{PR} &=(1\cdot2-(-1)\cdot2,\;(-1)(-1)-2\cdot2,\;2\cdot2-1\cdot(-1))\\ &=(2+2,\;1-4,\;4+1)\\ &=(4,-3,5). \end{aligned}

This nonzero vector is a normal. Using \(P\), the point-normal equation is \(4(x-1)-3(y-0)+5(z-2)=0\). Expanding gives \(4x-4-3y+5z-10=0\), or \(4x-3y+5z-14=0\).

Check all three points in the general form. For \(P\), \(4(1)-3(0)+5(2)-14=4+10-14=0\). For \(Q\), \(4(3)-3(1)+5(1)-14=12-3+5-14=0\). For \(R\), \(4(0)-3(2)+5(4)-14=0-6+20-14=0\). Each point satisfies the equation.

Worked Example: Simplifying the Normal Vector

Find the plane through \(P=(2,-1,0)\), \(Q=(4,0,1)\), and \(R=(1,2,3)\).

The direction vectors from \(P\) are

\overrightarrow{PQ}=(4-2,\;0-(-1),\;1-0)=(2,1,1), \qquad \overrightarrow{PR}=(1-2,\;2-(-1),\;3-0)=(-1,3,3).

The cross product is

\begin{aligned} \overrightarrow{PQ}\times\overrightarrow{PR} &=(1\cdot3-1\cdot3,\;1\cdot(-1)-2\cdot3,\;2\cdot3-1\cdot(-1))\\ &=(3-3,\;-1-6,\;6+1)\\ &=(0,-7,7). \end{aligned}

We may divide this normal by \(7\), obtaining the simpler normal \((0,-1,1)\). Using \(P\), the equation is \(0(x-2)-(y+1)+z=0\), which simplifies to \(z-y-1=0\).

Check \(P\): \(0-(-1)-1=0\). Check \(Q\): \(1-0-1=0\). Check \(R\): \(3-2-1=0\). All three points lie on the plane described by the equation.

Worked Example: A Plane with No \(y\)-Term

Find the plane through \(P=(0,1,2)\), \(Q=(2,1,0)\), and \(R=(1,3,1)\).

First form the directions:

\overrightarrow{PQ}=(2-0,\;1-1,\;0-2)=(2,0,-2), \qquad \overrightarrow{PR}=(1-0,\;3-1,\;1-2)=(1,2,-1).

Compute the cross product component by component:

\begin{aligned} \overrightarrow{PQ}\times\overrightarrow{PR} &=(0\cdot(-1)-(-2)\cdot2,\;(-2)\cdot1-2\cdot(-1),\;2\cdot2-0\cdot1)\\ &=(0+4,\;-2+2,\;4-0)\\ &=(4,0,4). \end{aligned}

Using the simpler normal \((1,0,1)\) and point \(P\), we obtain \(1(x-0)+0(y-1)+1(z-2)=0\), so the plane equation is \(x+z-2=0\).

Check \(P\): \(0+2-2=0\). Check \(Q\): \(2+0-2=0\). Check \(R\): \(1+1-2=0\). The equation is verified for each point.

A Common Mistake: Collinear Points

A cross product of zero is not a usable normal vector. It signals that the directions used to construct it are parallel, so the three points are collinear and do not specify one unique plane.

Worked Example: When the Points Do Not Determine a Unique Plane

Consider \(P=(1,2,0)\), \(Q=(3,5,1)\), and \(R=(5,8,2)\). The direction vectors from \(P\) are

\overrightarrow{PQ}=(2,3,1), \qquad \overrightarrow{PR}=(4,6,2).

Since \(\overrightarrow{PR}=2\overrightarrow{PQ}\), the vectors are parallel. Their cross product is therefore

\overrightarrow{PQ}\times\overrightarrow{PR} =(3\cdot2-1\cdot6,\;1\cdot4-2\cdot2,\;2\cdot6-3\cdot4) =(0,0,0).

A mistaken approach would treat \((0,0,0)\) as a normal and substitute it into a point-normal equation. That gives \(0=0\), not a plane equation. The correction is to recognize that the points are collinear: they all lie on the line through \(P\) in direction \((2,3,1)\). Many planes can contain that line, so these three points do not determine a unique plane.

Reliable Checks

After finding an equation, substitute the coordinates of all three points into it. A single successful substitution is not enough: the equation must contain every given point. If one point fails the check, revisit the vector subtractions, the cross-product components, and the constant term in the equation.

The cross product can point in either of two opposite normal directions. Reversing the order of the two direction vectors changes \(\mathbf{n}\) to \(-\mathbf{n}\), which changes both sides of the point-normal equation by a factor of \(-1\). Thus the equation describes the same plane. A nonzero scalar multiple of a normal likewise gives an equivalent equation, so simplifying a normal vector is safe.

This construction converts geometric data—the three points—into a standard algebraic description. The essential checks are that the cross product is nonzero and that the final equation is satisfied by each point.

Check Your Understanding

Use the construction and verification method to answer these questions.

What two vectors would you form from three given points before taking a cross product?
Why is the cross product a normal vector to the plane through three noncollinear points?
If the cross product is zero, what does that tell you about the three points?
Why may a nonzero normal vector be multiplied by a nonzero scalar without changing the plane?
What substitutions should be made to verify a proposed equation for a plane through three points?

What You'll Learn