Solutions for selected practice problems for Midterm 312.Review.22.Denote that linex= 1 +t,y= 2-t,z=-1 + 2tbyL,the origin by ), and the point onLwhich is closest to the origin byP.Thevector~OPmust be perpendicular toL’s direction vector~v=h1,-1,2i. IfPcorresponds to the value of the parametert, then~OP=h1 +t,2-t,-1 + 2ti,and we have 0 =~v=~OP= (1 +t)-(2-t) + 2(-1 + 2t) = 6t-3. This yieldt= 1/2. Thus, the point onLnearest to the origin isP(3/2,3/2,0), and thedistance from the origin to the line is|OP|=3/√2.Alternative solution.The square of the distance from the origin to thepointx= 1+t,y= 2-t,z=-1+2tequalsf(t) = (1+t)2+(2-t)2+(-1+2t)2=6-6t+ 6t2. We have to minimize this function.12.Review.25.The planesx-z= 1 andy+ 2z= 3 are not parallel (lookat their normal vectors), hence their intersection is a lineL. The line is per-pendicular to their normal vectors~n1=h1,0,-1iand~n2=h0,1,2i, hence wecan assume the direction vector ofLis~v=~n1×~n2=h1,-2,1i. To determineL, we also need to find a point on it. SinceLis not parallel to the coordinateplanes, we can look at the intersection ofLwith thexy-plane. There,z= 0,hencex= 1 andy= 3. Hence, our planePmust pass through (1,3,0).What about the normal vector toP? It must be perpendicular to~v=h1,-2,1i,and to~n3=h1,1,-2i(normal to the planex+y-2z= 1). Thus, the vector~v×~n3=h1,-2,1i × h1,1,-2i=h3,3,3i. We can take~n=h1,1,1ias normalto our plane.This leads to an equation (x-1) + (y-3) + (z-0) = 0, orx+y+z= 4.12.Review.35.Completing the equation 4x2+ 4y2-8y+z2= 0 to a square,we obtain 4x2+ 4(y-1)2+z2= 22. This is anellipsoid.12.Review.36.x=y2+z2-2y-4z+ 5 = (y-1)2+ (z-2)2is a(circular)paraboloid.13.Review.21. (a)~v=~r0=~R+t~R0.(b)~a=~v0= 2~R0+t~R00.14.Review.44.We need to remember thatD~uf(~r0) =~u· ∇f(~r0).