Math 251, Fall 2015-copyright Joe Kahlig Page 11. False: Lines could be skew.False: only unit if the angle between the vectors is 90 degreesTrue.True.2. Hyperboloid of one sheet: x2− 4y2+ 2z2= 10ellipsoid: 2x2+ 2y2+ z2= 50circular Paraboloid: 5y = 30x2+ 30z2hyperbolic cylinder: 4y2− 6z2= 24cone: 2x2+ 4y2− z2= 03. complete the square:center (−3, 5, 0) and radius√484. Let the second vector be h0, 0, 1i. now use the dot product formula to get.cos γ =3√145. projca =a · c|c|2c =16,26,166. compute a scalar triple product of these vectors: a ·(b × c), to get that the volume is 226.7. The direction vector for the line is v1= h2, 1, 7i. A vector from the point (1, 3, 4) to the point(3, 3, 5) is v2= h2, 0, 1i.The normal vector for the plane is v1× v2= h1, 12, −2iAnswer: x + 12y − z = 298. r0(t) = h3 cos t, 4, −3 sin ti and |r0(t)| =√25 = 5(a) L =5R1|r0(t)|dt = 20(b) s =tR0|r0(u)|du =tR05du = 5tr(s) =3 sins5,4s5, 3 coss5(c) T (t) =r0(t)|r0(t)|=35cos t,45,−35sin t.(d) κ =|T0(t)||r0(t)|=3259. Since the line is perpendicular to the plane, this means the direction vector of the line is parallelto the normal vector of the plane. so let the direction vector be some scalar multiple of thenormal vector.v = n = h3, 2, 4iAnswer: x = 1 + 3t, y = 1 + 2t, z = 1 + 4t10. an equation of the line is x = 1 + 4t, y = −t, and z = 2. plug this into the formula for thesurface and solve for t.you get t = 1 and t = 3points: (5, −1, 2) and (13, −3, 2)Check the back of the page for more problems.Math 251, Fall 2015-copyright Joe Kahlig Page 211. t = 2 gives the point so the tangent vector(direction vector) is v = f0(2) = h2, 16, 12ix = 5 + 2t, y = 16 + 16t, z = 8 + 12t12. κ =|r0(t) × r00(t)||r0(t)|3=√8(√1 + 8t2)313. Method 1: If there is a plane, P2, containing the line, L, where P2is parallel to the given plane,P, then the normal vector of P and the normal vector of P2are parallel. Since the line lies in P2then the dot product of the direction vector of the line and the normal vector of P2will be zero.In similar fashion the dot product of the direction vector of L and plane P will also be zero. Thedirection vector of the line is v = h4, 2, −1i and normal vector of plane P is n = h1, 3, 1i.v · n = 9 6= 0.so the answer is no. there is not a plane containing line L that will be parallel to plane P.Method 2: If there is a plane, P2, containing the line, L, where P2is parallel to the given plane,P , then the line will not intersect the plane. So check to see if there is a value of t will the linewill intersect the plane.x + 3y + z = 231 + 4t + 3(1 + 2t) + 1 − t = 23lots of workt = 2so the line intersects the plane so it is not parallel to the plane.Check the back of the page for more
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