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# Unit_7 - GREENBERG VECTOR GEOMETRY UNIT 7 LINES AND PLANES...

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UNIT 7 LINES AND PLANES INTRO: This Unit is about lines and planes: their equations, how they are determined by points and vectors, and some additional facts about them. In the discussions of this Unit, x , y and z will be variables, and a , b , c and d will be constants. 1. Lines It is very surprising to students that the equation y = mx + b , or equivalently, ax + by = c , is not necessarily the equation of a line, that is to say, the graph of the equation may not be a line. It is all a question of dimensions! Certainly, in two dimensions (in R 2 ) the equation ax + by = c is the equation of a line. We also say that in the x-y plane it is the equation of a line. However, in three dimensions (in R 3 ) it is the equation of a plane. In fact, in three dimensions any equation of the form ax + by + cz = d is the equation of a plane, including, for example, the case when c = 0, in which case the equation becomes ax + by = d , which in two dimensions would have been a line, but in three dimensions is a plane. We will study the equation of a plane in the next section. What, then, is the equation of a line in three dimensions? It is not possible to describe a line in three dimensions by a single Cartesian equation, meaning a single equation just in terms of x , y and z (although it is possible to specify a line as the intersection of two Cartesian equations). For this reason we prefer to give equations of lines in three dimensions as parametric equations. The equation of a line through the point P = ( p 1 , p 2 , p 3 ) going in the direction of the vector ~v = < v 1 , v 2 , v 3 > is x = p 1 + v 1 t y = p 2 + v 2 t z = p 3 + v 3 t We will consider this problem, where the direction ~v and a single point P on the line are given, as the model problem . For such a problem, we can write down the parametric equations of the line immediately, just as above. All other problems will be turned into the model problem. Since at least one point on the line is specified in all problems, the key to any line problem will be TO FIND ~v !!! Some examples. To find the equation of the line through the points Q = ( q 1 , q 2 , q 3 ) and R = ( r 1 , r 2 , r 3 ), choose ~v = Q - R VECTOR GEOMETRY 85 GREENBERG

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or, equivalently, ~v = R - Q , and either of the given points as your point P . To find the line through the point P = ( p 1 , p 2 , p 3 ) parallel to the line x = d + at, y = e + bt, z = f + ct , realizing that the coefficients of t give the vector ~v , use ~v = < a, b, c > and of course the specified point. To find the line through the point P = ( p 1 , p 2 , p 3 ) perpendicular to the plane ax + by + cz = d , we need a fact proved in the next section: the vector ~n = < a, b, c > is perpendicular to the plane. But this is just the direction we need, so use ~v = < a, b, c > and again the specified point. Example : All examples use the same strategy – find a vector ~v in the direction of the line, and then use it and one point to write the equation of the line. For example, to find the line through P = (1 , 2 , 3) parallel to the line x = 4 + 3 t y = 1 + t z = 2 - t.
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