POSITION_VECTOR said:
first of all, I'm sorry to have yelled at you...didn't mean it...especially to a professor.
Yelling? I didn't notice any yelling- I thought you were being very polite! You should see me when I yell!
to describe a plane, one uses two position vectors to locate two points on the plane. that line that forms on the plane must be used to form the dot product with the normal vector of that line. yes...it will be zero.
Still not clear what you mean.
Two points and two position vectors do
not determine a plane. One line in the plane is not enough to determine a plane (imagine rotating the plane around the given line). A single line has an infinite number of (unit) normal vector. There is a
plane normal to the line and any vector in that plane is a normal vector to the line.
here are the position vectors.
ro = <xo, yo, zo>
r = <x, y, z>
the line between the ends of these two points form on the plane we want to describe. let's call that vector p. let's say that p has the same components as the direction vector, v.
v = <a, b, c> = p
p = <at, bt, ct> and in this case...t =1 because we said that p = v.
ro + p = r
ro + v = r
<xo + at, yo +bt, zo +ct> = <x, y, z>
these are your parametric equations.
Yes, those are parametric equations for the
line between your two points (strictly speaking the "parametric equations" are x= x0+ at, y= y0+ bt, z= z0+ ct. What you give is a vector equation for the line) NOT equations for a plane which is what I thought you were saying.
because we said that v = p
<xo + a, yo + b, zo + c> = <x, y, z>
now...we do the dot product to find the LINEAR EQUATION OF THE PLANE.
p = r - ro
n = normal vector = <h, i, j>
n dot p = 0
n dot (r - ro) = 0
that becomes hx + iy + jz = h(xo) + i(yo) + j(zo)
which simplifies to
h(x - xo) + i(y - yo) + j(z - zo) = 0
this is where i am stuck...i'm not very sure what happened to D. if you find out what happened to D...is it the parameter?
Thanks for the help
Okay, now I see what you are doing. However, you are
not just using just two points in the plane as you said before. You are assuming that you are given two points in the plane
and the normal vector to the plane.
Actually, as I said before, if you are given the normal vector, the
one point is sufficient. The "vector" you need in the plane is between the given point x_0i+ y_0j+ z_0k and the "general point" xi+ yj+ zk where x, y, z are variables. Then your "r" is r= (x-x_0)i+ (y-y_0)j+ (z-z_0)k. With normal vector Ai+ Bj+ Ck.<br />
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(you are confusing the issue by writing the normal vector as "<h, i, j>= hx+ jy+ kz" where you seem to be using x, y, z as basis vectors. That is not standard notation and if you do that, you <b>cannot</b> then use x, y, z as variables. Standard notation is the x, y, z represent coordinates of a variable point in the plane and the unit vectors in the x, y, z directions are i, j, k respectively.)<br />
<br />
Writing the normal vector as <A, B, C>= Ai+ Bj+ Ck, the dot product becomes &lt;A, B, C&gt; \cdot &lt;x-x_0, y-y_0,z-z_0&gt; or, more formally (Ai+ Bj+ Ck)\cdot (x-x_0)i+ (y-y_0)j+ (z-z_0)k both of which equal A(x-x_0)+ B(y-y_0)+ C(z-z_0). Since the two vectors are perpendicular, the product is 0: A(x-x_0)+ B(y-y_0)+ C(z-z_0)= 0.&lt;br /&gt;
Multiplying that out gives Ax+ By+ Cz- Ax_0- By_0- Cz_0= 0 Which we can write as Ax+ By+ Cz= Ax_0+ By_0+ Cz_0. Since A, B, C, x&lt;sub&gt;0&lt;/sub&gt;, y&lt;sub&gt;0&lt;/sub&gt;, z&lt;sub&gt;0&lt;/sub&gt; are simply numbers (as opposed to the variables x, y, z) Ax&lt;sub&gt;0&lt;/sub&gt;+ By&lt;sub&gt;0&lt;sub&gt;+ Cz&lt;sub&gt;0&lt;/sub&gt; is simply a number. Call it &amp;quot;D&amp;quot; and you get Ax+ By+ Cz= D. No, D is &lt;b&gt;not&lt;/b&gt; a parameter- it is a fixed numbers and does not vary. The equation Ax+ By+ Cz= D is &lt;b&gt;not&lt;/b&gt; a parametric equation. &lt;br /&gt;
&lt;br /&gt;
Example: write the equation for the plane having normal vector 3i+ 2j+ 4z= &amp;lt;3, 2, 4&amp;gt; and containing the point (1, 1, 3). If (x, y, z) is a variable point in the plane, then (x-1)i+ (y-1)j+ (z-3)k= &amp;lt;x-1, y-1, z-1&amp;gt; is a vector in the plane. The dot product with the normal vector is&lt;br /&gt;
(3i+ 2j+ 4k)\cdot((x-1)i+ (y-1)j+ (z-3)k)= 3(x-1)+ 2(y-1)+ 4(z-3)= 0. That, of course, is the same as 3x- 3+ 2y- 1+ 4z- 3= 3x+ 2y+ 4z -(3+ 1+ 3)= 0 or 3x+ 2y+ 4z= 7. &amp;quot;D&amp;quot; is just the number 7 and is &lt;b&gt;not&lt;/b&gt; a parameter. This is not a &amp;quot;parametric&amp;quot; equation.&lt;br /&gt;
&lt;br /&gt;
You &lt;b&gt;can&lt;/b&gt; write this plane in terms of parametric equations but since a plane is 2 dimensional, you need two parameters. One easy way (there are always an infinite number of parametric equations for a surface depending on choice of parameter) is to take x and y themselves as parameter. That is, write x= u, y= v and the the equation above for the plane becomes 3u+ 2v+ 4z= 7 so 4z= 7- 3u- 2v and the parametric equations are x= u, y= v, z= (7/4)- (3/4)u- (1/2)v.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Parametric equations&amp;quot; are equations of the form &amp;quot;x= f(u,v), y= g(u,v), z= h(u,v)&amp;quot;- a different equation for each coordinate, depending (in the two dimensional case) on two parameters (parametric equations of a line or curve depend on one parameter). An equation of the form Ax+ By+ Cz= D is &lt;b&gt;not&lt;/b&gt; a &amp;quot;parametric equation&amp;quot; and has no parameter.&lt;/sub&gt;&lt;/sub&gt;
