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 $\Pi$ . For example, $\vec n=(1,1,5)$ .

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

This entry was posted on Tuesday, September 2nd, 2008 at 5:57 pm and is filed under 275: Calculus III. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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

William Foster says:

September 3, 2008 at 11:20 pm

Have we gotten our Homework 2 assigned yet?

Reply
andrescaicedo says:

September 4, 2008 at 6:31 pm

Hi William,

HW 2 is posted now under syllabus.

Reply
William Foster says:

September 4, 2008 at 9:09 pm

Thanks a bunch

Reply

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