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 u\cdot\vec 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}\cdot\vec n=-2+1+5=4 and \|\vec n\|=\sqrt{1+1+25}=\sqrt{27}, so \|\vec w\|=\frac4{3\sqrt3}.

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 \frac\pi2 -\theta. (Drawing a diagram may help you clarify the situation.)

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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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