275- Distance from a point to a plane

In class we tried to find the distance from P=(1,1,1) to the plane Pi of equation x+y+5z=3.

There are several ways of doing this. For example:

  • Fix a point on the plane Pi. Any point will do, say Q=(3,0,0).
  • Find a vector perpendicular to the Pi. For example, vec n=(1,1,5).

[Again, if a plane has equation Ax+By+Cz=D then vec n=(A,B,C) is perpendiculart to it.

For example, x+y+5z=3 is parallel to the plane x+y+5z=0, which we can rewrite as (1,1,5)cdot(x,y,z)=0, which is the equation of the set of points (x,y,z) perpendicular to the vector (1,1,5). This is to say, the plane x+y+5z=0 is perpendicular to the vector (1,1,5). Since the plane Pi: x+y+5z=3 is parallel to x+y+5z=0, Pi is also perpendicular to (1,1,5).

Another way of reaching the same conclusion is to rewrite Ax+By+Cz=D in the form A(x-x_0)+B(y-y_0)+C(z-z_0)=0 for some appropriate vector (x_0,y_0,z_0). There are many choices of (x_0,y_0,z_0) and they all work; all we need is that Ax_0+By_0+Cz_0=D, i.e., that (x_0,y_0,z_0) is in the original plane. For example, the point (3,0,0) belongs to the plane Pi: x+y+5z=3, so we can rewrite the equation of Pi as (x-3)+y+5z=0, which is equivalent to saying that (1,1,5)cdot[(x,y,z)-(3,0,0)]=0. null But this means that the vector vec n=(1,1,5) is perpendicular to the vector (x,y,z)-(3,0,0), which is an arbitrary vector in the direction of the plane.]

Let’s continue with the problem of finding the distance from P to Pi:

  • Consider the projection vec w={rm proj}_{vec n}overrightarrow{QP} of the vector overrightarrow{QP} in the direction of vec n. Clearly, the distance from P to Pi is the length of vec w.
  • Recall that {rm proj}_{vec u}vec v=frac{vec ucdotvec v}{|vec u|^2}vec u.
  • Then vec w=frac{overrightarrow {QP} cdot vec n}{|vec n|^2}vec n and the distance is |vec w|=frac{|overrightarrow {QP} cdot vec n|}{|vec n|}.
  • In detail, overrightarrow{QP}=(-2,1,1) so overrightarrow{QP}cdotvec n=-2+1+5=4 and |vec n|=sqrt{1+1+25}=sqrt{27}, so |vec w|=frac4{3sqrt3}.

Notice that (as discussed in class) the distance from a point P to the line in the direction of vec v that goes through a point Q is given by d=frac{|overrightarrow{QP}times vec v|}{|vec v|}, while (by the above) the distance from a point P to a plane containing a point Q and perpendicular to a vector vec v is given by d=frac{|overrightarrow {QP} cdot vec v|}{|vec v|}. While the expressions are similar, one involves a cross product and the other a dot product. This is because in one case we express the distance in terms of the sine of an angle theta, and in the other, in terms of its cosine or, what is the same, in terms of the sine of fracpi2 -theta. (Drawing a diagram may help you clarify the situation.)


3 Responses to 275- Distance from a point to a plane

  1. William Foster says:

    Have we gotten our Homework 2 assigned yet?

  2. Hi William,

    HW 2 is posted now under syllabus.

  3. William Foster says:

    Thanks a bunch

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