Drawing a plane through three vectors.

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November 28th 2011, 05:27 AM
shtonki

Drawing a plane through three vectors.

I wan't to start off by saying that I am not taking the course in English, so if I use poorly translated terms, please ask me to clarify.

I have three vectors P,Q,R, and I have been asked to:

1. Calculate the plane that goes through the three vectiors, in normal form.

2. Calculate the shortest distance from origo to the plane.
November 28th 2011, 05:46 AM
HallsofIvy

Re: Drawing a plane through three vectors.

In general, given three vectors, there does not exist a plane containing those three vectors. So I suspect you P, Q, R are actually three points, or the position vectors of those three points, and you want to find the plane containing those three points. If your points are given by $P= (p_1, p_2, p_3)$ (equivalently vector $P= p_1\vec{i}+ p_2\vec{j}+ p_3\vec{k}$ ), $Q= (q_1, q_2, q_3)$ ( $Q= q_1\vec{i}+ q_2\vec{j}+ q_3\vec{k}$ ), and $R= (r_1, r_2, r_3)$ ( $R= r_1\vec{i}+ r_2\vec{j}+ r_3\vec{k}$ ) then vectors in the plane are given by the difference of two of those. In particular, $U= P- Q= (p_1- q_1)\vec{i}+ (p_2- q_2)\vec{j}+ (p_3- q_3)\vec{k}$ and $V= R- Q= (r_1- q_1)\vec{i}+ (r_2- q_2)\vec{j}+ (r_3- q_3)\vec{k}$ are vectors in the plane. Their cross product will give a vector, $A\vec{i}+ B\vec{j}+ C\vec{k}$ , perpendicular to the plane. And the equation of a plane having that normal vector, containing point $(x_0, y_0, z_0)$ is given by $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$ .

The shortest distance from a point to a plane (or line) is along a perpendicular line. Once you have the plane as $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$ , you know that $A\vec{i}+ B\vec{j}+ C\vec{k}$ is perpendicular to the plane and so x= At, y= Bt, z= Ct is a line perpendicular to the plane, through (0, 0, 0). Replace x, y, and z in the equation of the plane and you have a single equation to solve for t, giving the point where that line hits the plane. The distance between that point and (0, 0, 0) is the shortest distance from (0, 0, 0) to the plane.

(If you are doing this for a course, I bet all of that is in your textbook!)
November 28th 2011, 06:16 AM
shtonki

Re: Drawing a plane through three vectors.

Big thanks for taking the time to help me, would you mind making sure i have understood how to use this?

So in my example i have P = {1,3,7}, Q = {2, 0, 4} and R = {5,4,1}.

I first want to calculate U = P-Q = {-1, 3, 3}
V = R-Q = {3, 4, -3}

UxV = {21, 6, -13} which is the plane's normal vector, gives us

21(x-1) + 6(y-3) - 13(z-7) = 0 ?
November 28th 2011, 06:27 AM
Plato

Re: Drawing a plane through three vectors.

Quote:

Originally Posted by shtonki

So in my example i have P = {1,3,7}, Q = {2, 0, 4} and R = {5,4,1}.
I first want to calculate U = P-Q = {-1, 3, 3} & V = R-Q = {3, 4, -3}
UxV = {21, 6, -13} which is the plane's normal vector, gives us
21(x-1) + 6(y-3) - 13(z-7) = 0 ?

I have not actually checked the numbers, but the method is correct.

Here is a note on notation.
It is customary to write a point as $(1,3,7)$ , whereas a vector as $<21,6,-13>$ .
November 28th 2011, 06:56 AM
shtonki

Re: Drawing a plane through three vectors.

Thanks, I'll keep that in mind.
November 28th 2011, 07:56 AM
Hartlw

Re: Drawing a plane through three vectors.

In the vector spirit:

n = (q-p)X(r-p)/(l(q-p)X(r-p)l)

d = p.n

x.n = d

p, q, r vectors to points in space.
n normal to plane through the points.
d distance from origin to plane.
x vector to a point in the plane.
X and . cross and dot product.

EDIT: Used wrong magnitude to norm n. Corrected it.