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Mathematics LibreTexts

4.8: Planes in Rⁿ

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Outcomes

  1. Find the vector and scalar equations of a plane.

Much like the above discussion with lines, vectors can be used to determine planes in Rn. Given a vector n in Rn and a point P0, it is possible to find a unique plane which contains P0 and is perpendicular to the given vector.

Definition 4.8.1: Normal Vector

Let n be a nonzero vector in Rn. Then n is called a normal vector to a plane if and only if nv=0 for every vector v in the plane.

In other words, we say that n is orthogonal (perpendicular) to every vector in the plane.

Consider now a plane with normal vector given by n, and containing a point P0. Notice that this plane is unique. If P is an arbitrary point on this plane, then by definition the normal vector is orthogonal to the vector between P0 and P. Letting 0P and 0P0 be the position vectors of points P and P0 respectively, it follows that n(0P0P0)=0 or nP0P=0

The first of these equations gives the vector equation of the plane.

Definition 4.8.2: Vector Equation of a Plane

Let n be the normal vector for a plane which contains a point P0. If P is an arbitrary point on this plane, then the vector equation of the plane is given by n(0P0P0)=0

Notice that this equation can be used to determine if a point P is contained in a certain plane.

Example 4.8.1: A Point in a Plane

Let n=[123] be the normal vector for a plane which contains the point P0=(2,1,4). Determine if the point P=(5,4,1) is contained in this plane.

Solution

By Definition 4.8.2, P is a point in the plane if it satisfies the equation n(0P0P0)=0

Given the above n, P0, and P, this equation becomes [123]([541][214])=[123]([333])=3+69=0

Therefore P=(5,4,1) is contained in the plane.

Suppose n=[abc], P=(x,y,z) and P0=(x0,y0,z0).

Then n(0P0P0)=0[abc]([xyz][x0y0z0])=0[abc][xx0yy0zz0]=0a(xx0)+b(yy0)+c(zz0)=0

We can also write this equation as ax+by+cz=ax0+by0+cz0

Notice that since P0 is given, ax0+by0+cz0 is a known scalar, which we can call d. This equation becomes ax+by+cz=d

Definition 4.8.3: Scalar Equation of a Plane

Let n=[abc] be the normal vector for a plane which contains the point P0=(x0,y0,z0).Then if P=(x,y,z) is an arbitrary point on the plane, the scalar equation of the plane is given by ax+by+cz=d where a,b,c,dR and d=ax0+by0+cz0.

Consider the following equation.

Example 4.8.2: Finding the Equation of a Plane

Find an equation of the plane containing P0=(3,2,5) and orthogonal to n=[241].

Solution

The above vector n is the normal vector for this plane. Using Definition 4.8.2, we can determine the vector equation for this plane. n(0P0P0)=0[241]([xyz][325])=0[241][x3y+2z5]=0

Using Definition 4.8.3, we can determine the scalar equation of the plane. 2x+4y+1z=2(3)+4(2)+1(5)=9

Hence, the vector equation of the plane is [241][x3y+2z5]=0 and the scalar equation is 2x+4y+1z=9

Suppose a point P is not contained in a given plane. We are then interested in the shortest distance from that point P to the given plane. Consider the following example.

Example 4.8.3: Shortest Distance From a Point to a Plane

Find the shortest distance from the point P=(3,2,3) to the plane given by
2x+y+2z=2, and find the point Q on the plane that is closest to P.

Solution

Pick an arbitrary point P0 on the plane. Then, it follows that QP=projnP0P and QP is the shortest distance from P to the plane. Further, the vector 0Q=0PQP gives the necessary point Q.

From the above scalar equation, we have that n=[212]. Now, choose P0=(1,0,0) so that n0P=2=d. Then, P0P=[323][100]=[223].

Next, compute QP=projnP0P. QP=projnP0P=(P0Pnn2)n=129[212]=43[212]

Then, QP=4 so the shortest distance from P to the plane is 4.

Next, to find the point Q on the plane which is closest to P we have 0Q=0PQP=[323]43[212]=13[121]

Therefore, Q=(13,23,13).


This page titled 4.8: Planes in Rⁿ is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform.

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