Solutions to old Exam 1 problemsHi students!I am putting this old version of my review for the first midtermreview, place and time to be announced. Check for updates onthe web site as to which sections of the book will actually becovered.Enjoy!!Best,Bill MeeksPS. There are probably errors in some of the solutionspresented here and for a few problems you need to completethem or simplify the answers; some questions are left to youthe student. Also you might need to add more detailedexplanations or justifications on the actual similar problems onyour exam. I will keep updating these solutions with bettercorrected/improved versions.Problem 1(a) - Fall 2008Findparametric equationsfor the lineLwhich containsA(1,2,3)andB(4,6,5).Solution:To get theparametric equationsofLyou need a pointthrough which the line passes and a vector parallel to the line.Take the point to beAand the vector to be the−→AB.The vector equation ofLisr(t) =−→OA+t−→AB=⟨1,2,3⟩+t⟨3,4,2⟩=⟨1 + 3t,2 + 4t,3 + 2t⟩,whereOis the origin.Theparametric equationsare:x= 1 + 3ty= 2 + 4t,t∈R.z= 3 + 2tProblem 1(b) - Fall 2008Findparametric equationsfor the lineLof intersection of theplanesx−2y+z= 10 and 2x+y−z= 0.Solution:The vector partvof the lineLof intersection is orthogonal tothe normal vectors⟨1,−2,1⟩and⟨2,1,−1⟩. Hencevcan betaken to be:v=⟨1,−2,1⟩ × ⟨2,1,−1⟩=ijk1−2121−1= 1i+ 3j+ 5k.ChooseP∈Lso thez-coordinate ofPis zero. Settingz= 0,we obtain:x−2y= 102x+y= 0.Solving, we find thatx= 2 andy=−4.Hence,P=⟨2,−4,0⟩lies on the lineL.Theparametric equationsare:x= 2 +ty=−4 + 3tz= 0 + 5t= 5t.Problem 2(a) - Fall 2008Find anequation of the planewhich contains the pointsP(−1,0,1),Q(1,−2,1) andR(2,0,−1).Solution:Method 1Consider the vectors−→PQ=⟨2,−2,0⟩and−→PR=⟨3,0,−2⟩which lie parallel to the plane.Then consider the normal vector:n=−→PQ×−→PR=ijk2−2030−2= 4i+ 4j+ 6k.So theequation of the planeis given by:⟨4,4,6⟩·⟨x+ 1,y,z−1⟩= 4(x+ 1) + 4y+ 6(z−1) = 0.Problem 2(a) - Fall 2008Find anequation of the planewhich contains the pointsP(−1,0,1),Q(1,−2,1) andR(2,0,−1).Solution:Method 2The plane consists of all the pointsS(x,y,z)∈R3, such that−→PS,−→PQand−→PRare in the same plane (coplanar).But this happens if and only if their box product is zero.So theequation of the planeis:x+ 1yz−12−2030−2= 4(x+ 1) + 4y+ 6(z−1) = 0.Problem 2(b) - Fall 2008Find the distanceDfrom the point (1,6,−1) to the plane2x+y−2z= 19.Solution:Recall the distance formulaD=|ax1+by1+cz1+d|√a2+b2+c2from a pointP= (x1,y1,z1) to a planeax+by+cz+d= 0.