Math 53 Worksheet Solutions - Cross Products and Planes1. Prove that (a − b) × (a + b) = 2(a × b).Solution.(a − b) × (a + b) = a × (a + b) − b × (a + b)= a × a + a × b − b × a − b × b= a × b + a × b= 2(a × b)2. Use the scalar triple product to determine whether the points A(1, 3, 2), B(3, −1, 6),C(5, 2, 0), and D(3, 6, −4) lie in the same plane.Solution. Set a = AB = h2, −4, 4i, b = AC = h4, −1, −2i, and c = AD = h2, 3, −6i.We need to know if this vectors lie in the same plane. As such, we take the scalartriple product.a · (b × c) =2 −4 44 −1 −22 3 −6= 0.As the scalar triple product is zero, they are coplanar.3. Find the distance from the point (1, −2, 4) to the plane 3x + 2y + 6z = 5.Solution. Use the formula from the book. a = 3, b = 2, c = 6, d = −5, and(x1, y1, z1) = (1, −2, 4). SoD =|3(1) + 2(−2) + 6(4) − 5|√32+ 22+ 62=187.4. Give a geometric description of each family of planes.(a) x + y + z = c(b) x + y + cz = 1(c) y cos θ + z sin θ = 1Solution.(a) Normal vector is n = h1, 1, 1i. x, y, and z intercepts are all c. Draw a picture.Plane makes equilateral triangle in first octant (c > 0) or octant opposite the first(c < 0).(b) Normal vector is n = h1, 1, ci. x an y intercepts are 1, but z intercept is1c. Assuch, as c gets large, plane approaches the xy plane.1(c) Normal vector is n = h0, cos θ, sin θi. Note that an obvious point on the plane is(x, cos θ, sin θ) for any x. It follows that the family consists of planes tangent tothe cylinder with radius 1 and with the x axis as its major axis.5. If a, b, and c are all not 0, show that the equation ax + by + cz + d = 0 represents aplane and ha, b, ci is a normal vector to the plane.Solution. Assume a 6= 0. The proof for the other cases is similar. Then we cansimply move d into the x term, as inax −da+ by + cz = 0,and this now has the formha, b, ci · hx −da, y, zi = 0,which is the vector form of a plane.6. Find the distance between the skew lines with parametric equations x = 1 + t,y = 1 + 6t, z = 2t, and x = 1 + 2s, y = 5 + 15s, z = −2 + 6s.Solution. The direction vectors of the lines are v1= h1, 6, 2i and v2= h2, 15, 6i. Son = v1× v2should be perpendicular to both lines. We calculaten = v1× v2= h6, −2, 3i.Draw a picture of what this looks like. Now pick points on the lines, P1= (1, 1, 0) andP2= (1, 5, −2) will do. (Get this by setting t = s = 0.) Form the vector connectingthese points on the linesb = h0, 4, −2i.Then, by a geometry argument or by scalar projection, it follows thatD =|b · n||n|=14p62+ (−2)2+ 32=147= 2.7. Find equations of the planes that are parallel to the plane x + 2y − 2z = 1 and twounits away from it.Solution. We could use the next problem to do this easily, but there is an easier way.The normal vector to the plane is n = h1, 2, −2i. A vector in the same direction as thenormal vector but with length 2 (check this) isv =n|n|(2) = h23,43, −43i.2A point on the plane is P0= (1, 1, 1). The planes that are parallel and two units awaywill have identical normal vectors, but different points. Moving 2 units in the direction(or antidirection) of the normal vector from point P0will take us to new pointsP1= P0+ v = (53,73, −13), P2= P0− v = (13, −13,73).The new equations of the planes are then(1)(x −53) + 2(y −73) + (−2)(z +13) = 0, (1)(x −13) + (2)(y +13) + (−2)(z −73) = 0.These can be rewrittenx + 2y − 2z = 7, x + 2y − 2z = −5.8. Show that the distance between the parallel planes ax + by + cz + d1= 0 andax + by + cz + d2= 0 isD =|d1− d2|√a2+ b2+ c2.Solution. Pick a point on the second plane, and call it P1= (x1, y1, z1). Thenax1+ by1+ cz1+ d2= 0 and in factax1+ by1+ cz1= −d2.The distance between the first plane and this point is given by the formula in thebook, soD =|ax1+ by1+ cz1+ d1|√a2+ b2+ c2=| − d2+ d1|√a2+ b2+ c2,and this is the formula.9. If v1, v2, and v3are noncoplanar vectors, letk1=v2× v3v1· (v2× v3), k2=v3× v1v1· (v2× v3)k3=v1× v2v1· (v2× v3)(These vectors occur in the study of crystallography. Vectors of the normn1v1+ n2v2+ n3v3, where each niis an integer, form a lattice for a crystal. Vectorswritten similarly in terms of k1, k2, k3form the reciprocal lattice. )(a) Show that kiis perpendicular to vjif i 6= j.(b) Show that ki· vi= 1 for i = 1, 2, 3.(c) Show that k1· (k2× k3) =1v1·(v2×v3).Solution.3(a) For any vectors a and b the scalar triple productsa · (a × b) = 0, a · (b × a) = 0,so it follows that kiis perpendicular to vjwhen i 6= j.(b) Evaluate each one and use part (5) of Theorem 8.(c) Perform the calculation using the properties of the cross product in Theorem