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LESSON 3 Planes and Quadric Surfaces READ: Sections 13.5, 13.6 NOTES: Equations of planes in 3-space play an important role in vector calculus. Recall that one way to specify a line in the x , y -plane is to name a point on the line and a slope for the line. From those two items, an equation for the line can be determined. A plane in 3-space can also be speciﬁed by giving two pieces of information about the plane. One is a vector in a direction perpendicular to the plane, say -→ n = h a,b,c i , and the second item is a single point, P 0 = ( x 0 ,y 0 ,z 0 ), on the plane. If you picture things in your mind, you will see that when the vector -→ n has been speciﬁed, there will be a stack of planes to which that vector is perpendicular, sort of like sheets of paper sandwiched together in a pad. The given point that the plane must go through selects one particular sheet from the pad. The vector -→ n is called a normal vector for the plane. Notice that the magnitude of -→ n is not important; its only job is to determine a direction. If the normal vector is multiplied by any nonzero scalar, the new vector serves just as well as a normal vector for the plane. Often a normal vector is divided by its length to produce a unit normal vector since that makes many formulas look a little neater. Notice that there are just two unit normals to a plane, and they are negatives of each other. The particular point speciﬁed on the plane also does not matter. Any point on the plane can be used. The description of a plane via a normal vector and a point on the plane can be translated into a variety of equations that are satisﬁed by points P = ( x,y,x ) on the plane and no other points. Vector form: Using the notation above, let’s use P 0 for the point ( x 0 ,y 0 ,z 0 ), and P for the point ( x,y,z ). Since the point P 0 is on the plane, we see that P is on the plane if and only if the line segment from P 0 to P is in the plane, and that’s correct if and only if -→ n is perpendicular to that line segment, and that’s so if and only if -→ n is perpendicular to a vector in the direction of that line segment. Since --→ P 0 P = h x - x 0 ,y - y 0 ,z - z 0 i is a vector in the direction of that line segment, it follows that the point P = ( x,y,z ) is on the plane if and only if -→ n · --→ P 0 P = 0, since that dot product tests to see if the two vectors are perpendicular. In other words, P is on the plane if and only if -→ n · --→ P 0 P = 0 or -→ n · -→ P = -→ n · -→ P 0 . Either of those equations is called a

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