UNIT 7LINES AND PLANESINTRO: This Unit is about lines and planes: their equations, how they are determined bypoints and vectors, and some additional facts about them. In the discussions of this Unit,x,yandzwill be variables, anda,b,canddwill be constants.1. LinesIt is very surprising to students that the equationy=mx+b, or equivalently,ax+by=c, isnot necessarily the equation of a line, that is to say, the graph of the equation may not be aline. It is all a question of dimensions!Certainly, in two dimensions (inR2) the equationax+by=cis the equation of a line. We alsosay that in the x-y plane it is the equation of a line. However, in three dimensions (inR3) it isthe equation of a plane. In fact, in three dimensions any equation of the formax+by+cz=dis the equation of a plane, including, for example, the case whenc= 0, in which case theequation becomesax+by=d, whichin two dimensionswould have been a line, but in threedimensions 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 inthree dimensions by a single Cartesian equation, meaning a single equation just in terms ofx,yandz(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 pointP= (p1, p2, p3) going in the direction of the vector~v=< v1, v2, v3>isx=p1+v1ty=p2+v2tz=p3+v3tWe will consider this problem, where the direction~vand a single pointPon the line are given,as themodel problem. For such a problem, we can write down the parametric equations of theline immediately, just as above.All other problems will be turned into the model problem. Since at least one point on the lineis 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 pointsQ= (q1, q2, q3) andR= (r1, r2, r3), choose~v=Q-R

or, equivalently,~v=R-Q, and either of the given points as your pointP.To find the line through the pointP= (p1, p2, p3) parallel to the linex=d+at, y=e+bt, z=f+ct, realizing that the coefficients oftgive the vector~v, use~v=< a, b, c >and of course the specified point.To find the line through the pointP= (p1, p2, p3) perpendicular to the planeax+by+cz=d,we need a fact proved in the next section: the vector~n=< a, b, c >is perpendicular to theplane. 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~vin the direction of the line,and then use it and one point to write the equation of the line. For example, to find the linethroughP= (1,2,3) parallel to the linex= 4 + 3ty= 1 +tz= 2-t.one sees that~v=<3,1,-1>is the direction of the given line, whence it will also be used inthe answer, and the desired line isx= 1 + 3ty= 2 +tz= 3-tExample: Find the equations of a line through the pointsP= (3,-2,

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