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Planes

Defining a Plane using a Normal Vector and a Point

MathJax TeX Test Page A plane, simply defined, is an infinitely large flat surface. For example, the x-y plane is the 2d coordinate system. Now, to define a plane, it is written in the form: $$Ax + By + Cz = D$$ Let's say you have a normal vector: $ <1,2,5> $, and a point $ (-2, 7, 8) $.
Now, this is what we do:
First, we have to make up a point: $(x,y,z)$ which is where the normal vector intersects the plane.
Now, we draw the vector from $(-2, 7, 8)$ to $(x,y,z)$ = $< x + 2, y - 7, z - 8> $ and that must be perpendicular to the normal vector.
Now, recall from before, we took the dot products between the two vectors, and if they equalled zero, that meant that they were perpendicular. $$ < x+2,y-7,z-8> \cdot <1,2,5> = (x+2) + 2\cdot (y-7) + 5 \cdot (z-8) = 0$$ $$x + 2y + 5z + 2 - 14 - 40 = 0 \Rightarrow \boxed{x + 2y + 5z = 52} $$

Generalization

MathJax TeX Test Page Now, instead of using real values, let's generalize it. Let's say you have a normal vector $< A, B, C>$, which intersects the plane at $(x,y,z)$, and another point on the plane $(x_1, y_1, z_1)$. Let's just do the same thing. The vector connecting the two points is: $< x - x_1, y - y_1, z - z_1>$. By taking the dot product, we get: $$ A \cdot (x - x_1) + B \cdot (y - y_1) + C \cdot(z - z_1) = 0$$ $$ Ax + By + Cz = Ax_1 + By_1 + Cz_1$$ Now, what does this tell us? If you have a normal vector $< A,B,C>$ you can put it in the form of $Ax + By + Cz + D = 0$, and then all you have to do is plug in the point you know to get D.

Redo of the Initial Problem

MathJax TeX Test Page So, we have a normal vector $< 1,2,5>$, and a point $(-2, 7, 8)$ Right off the bat, we can write this: $$x + 2y + 5z = D$$ Now, we just plug in the point. $$-2 + 2 \cdot 7 + 5 \cdot 8 = D \Rightarrow D = 52$$ Here's our equation for the plane: $$x + 2y + 5z = 52$$

Defining a Plane using Three Points

MathJax TeX Test Page Let's say we have three points: $(1,-3, 5)$, $(4,5, 10)$, and $(0, 8, 6)$. In order to find the equation of the plane, we must turn these three points into the form we just learned: one point, and a normal vector. Now, as we learned before, a cross product is the vector that is perpendicular to two other vectors. In this case, we can make two vectors: We can make a vector from point 1 to 2, and another vector from point 1 to 3. $$\vec{v_1} = < 3,8,5>$$ $$\vec{v_2} = < -1,5,1>$$ Now, we just need to figure out the cross product. We learned how to do that before. That equals $< -17, -8, 23>$. We have three points to choose from, let's just choose the first one: $(1,-3,5)$.
We know how to solve this kind of problem. $$-17x - 8y + 23z = D$$ $$-17\cdot 1 - 8\cdot (-3) + 23 \cdot 5 = -17 + 24 + 115 = D = 122$$ $$\text{Our answer is } \boxed{-17x - 8y + 23z = 122}$$
David Witten

Distance between Objects in 3-Space

Define a Line using Vectors