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Calculus IIIMath 143 Spring 2008Professor Ben RichertExam 2SolutionsProblem 1.(20pts) Consider the pointsA= (1,2,3)andB= (2,3,5), and the line−⇀r(t) =(2,2,3)+t(1,2,5).(a–5pts) Give parametric equations for the line containingAand parallel to−−⇀AB.Solution.The line through (a, b, c) in direction(d, e, f)is(a, b, c)+t(d, e, f), giving parametric equationsx(t)=a+dty(t)=b+etz(t)=c+ftWe use the point (1,2,3) and the direction−−⇀AB=(1,1,2)which we get by subtracting the pointBfromA. So ourline is:x(t)=1 +ty(t)=2 +tz(t)=3 + 2tsquare(b–5pts) Give the equation of a plane which is perpendicular to−⇀r(t)and containsB.Solution.The equation of a plane containing (x0, y0, z0) with normal(a, b, c)isa(x−x0) +b(y−y0) +c(z−z0) = 0.Since the plane we are looking for is perpendicular to−−⇀AB, its normal must be parallel to−−⇀AB=(1,1,2), that is, wecan use(1,1,2)as its normal. With the pointB= (2,3,5), we get the plane(x−2) + (y−3) + 2(z−5) = 0.square(c–10pts) Give the equation of the plane containing−⇀r(t)and the pointB.Solution.LetC= (2,2,3), and note thatCis a point on−⇀r(t). Then the normal to our plane will be perpendicularto both(1,2,5), the direction of−⇀r(t), and−−⇀CB=(0,1,2). We know that the(1,2,5) × (0,1,2)=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle−⇀i−⇀j−⇀k125012vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=(−1,−2,1)is normal to both those directions. Thus the plane in question is−(x−2)−2(y