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The Dot and Cross Product I. Quiz II. Homework III. The Dot Product We define the dot product of two vectors v = ai + bj and w = ci + dj to be v @ w = ac + bd (In reality the @ sign should be a centered dot, but loading a gif takes time) Notice that the dot product of two vectors is a number and not a vector For 3 dimensional vectors, we define the dot product similarly: if v = ai + bj + ck and w = di + ej + fk then v @ w = ad + be + cf Examples: If v = 2i + 4j and w = i + 5j, then v@w = (2)(1) + (4)(5) = 22 The class will find the dot product of 2i + j - k and i + 2j III. The Angle Between Two Vectors We define the angle theta between two vectors v and w by the formula cos(theta) = (v@w)/(||v|| ||w||) so that v@w = ||v|| ||w|| cos(theta) Two vectors are called orthogonal if their angle is a right angle. We see that angles are orthogonal if and only if v@w = 0. Example To find the angle between v = 2i + 3j + k and w =4i + j + 2k we compute: ||v|| = sqrt(4 + 9 + 1) = sqrt14 ||w|| = sqrt(16 + 1 + 4) = sqrt(21) and v@w = 8 + 3 + 2 = 13 Hence theta = cos-1(13/sqrt[(14)(21)] IV. Direction Angles Definition Let v = ai + bj + ck be a vector, then we define the direction cosines to be the following: cos(alpha) = a/||v|| cos(beta) = b/||v|| cos(gamma) = c/||v|| V. Projections and Components Suppose that a car is stopped on a steep hill, and let g be the force of gravity acting on it. We can split the vector g into the component that is pushing the car down the road and the component that is pushing the car onto the road. We define Definition: Let u and v be a vectors. Then u can be broken up into two components, r and s such that r is parallel to v and s is perpendicular to v. r is called the projection of u onto v and s is called the component of u perpendicular to v. We see that u@v = ||u||||v||cos(theta) = ||u||||v||||s||/||u||= ||v||||s|| hence ||s|| = u@v/||v|| hence We can calculate the projection of u onto v by the formula: projvu = (u@v/||v||2)v VI. Work The work done by a constant force F along PQ is given by W = F@PQ III. Torque Suppose you are skiing and have a terrible fall. Your body spins around and you ski stays in place. With proper bindings your bindings will release and your ski will come off. The bindings recognize that a force has been applied. This force is called torque. To compute it we use the cross produce of two vectors which not only gives the torque, but also produces the direction that is perpendicular to both the force and the direction of the leg. IV. The Cross Product Between Two Vectors Definition: Let u = ai + bj + ck and v = di + ej + fk be vectors then we define the cross product v x w by the determinant of the matrix:
=
i -
j +
k = (bf - ce)i + (cd - af)j + (bf - ce)k If you need more help see the lecture notes for Math 103 B on matrices. The class will try the following examples: Find u x v when A) u = 3i + j - 2k, v = i - k B) u = 2i - 4j - k, v = 3i - j + 2k Notice that since switching the order of two rows of a determinant changing the sign of the determinant, we have u x v = -v x u V. Geometry and the Cross Product Let u and v be vectors and consider the parallelogram that the two vectors make. Then ||u x v|| = Area of the Parallelogram and the direction of u x v is a right angle to the parallelogram that follows the right hand rule (shown in class). Note that for i x j the magnitude is 1 and the direction is k, hence i x j = k. Exercise: Find j x k and i x k VI. Torque Revisited We define the torque (or the moment M of a force F about a point Q as M = PQ x F Example: A 20 inch wrench is at an angle of 30 degrees with the ground. A force of 40 pounds that makes and angle of 45 degrees with the wrench turns the wrench. Find the torque. Solution: We can write the wrench as the vector 20cos(30)i + 20sin(30)j = 17.3i + 10j and the force as -40cos(75)i - 40sin(75)j = -10.3i - 38.6j hence, the torque is the magnitude of their cross product:
= -564 foot pounds VII. Parallelepipeds To find the volume of the parallelepiped spanned by three vectors u, v, and w, we find the triple product: Volume = u@(v x w) = determinant of
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