4.8: Planes in Rⁿ
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- 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.
Let →n be a nonzero vector in Rn. Then →n is called a normal vector to a plane if and only if →n∙→v=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∙(→0P−→0P0)=0 or →n∙→P0P=0
The first of these equations gives the vector equation of the 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∙(→0P−→0P0)=0
Notice that this equation can be used to determine if a point P is contained in a certain 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∙(→0P−→0P0)=0
Given the above →n, P0, and P, this equation becomes [123]∙([541]−[214])=[123]∙([33−3])=3+6−9=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∙(→0P−→0P0)=0[abc]∙([xyz]−[x0y0z0])=0[abc]∙[x−x0y−y0z−z0]=0a(x−x0)+b(y−y0)+c(z−z0)=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
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,d∈R and d=ax0+by0+cz0.
Consider the following equation.
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∙(→0P−→0P0)=0[−241]∙([xyz]−[3−25])=0[−241]∙[x−3y+2z−5]=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]∙[x−3y+2z−5]=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.
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=proj→n→P0P and ‖→QP‖ is the shortest distance from P to the plane. Further, the vector →0Q=→0P−→QP gives the necessary point Q.
From the above scalar equation, we have that →n=[212]. Now, choose P0=(1,0,0) so that →n∙→0P=2=d. Then, →P0P=[323]−[100]=[223].
Next, compute →QP=proj→n→P0P. →QP=proj→n→P0P=(→P0P∙→n‖→n‖2)→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=→0P−→QP=[323]−43[212]=13[121]
Therefore, Q=(13,23,13).