Vectors Calaulus 1.2

The Dot and Cross Product The Dot Product                  Definition We define the dot product of two vectors      v = ai + bj  and w = ci + dj  to be           v . w = ac + bd  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: Dot Product in R3  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   Exercise Find the dot product of         2i + j - k     and     i + 2j The Angle Between Two Vectors We define the angle theta between two vectors v and w by the formula                             v . w                    cos q  =                                          ||v|| ||w|| so that          v . w = ||v|| ||w|| cos q 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:         and         and         v . w  =  8 + 3 + 2 = 13 Hence           Direction Angles      Definition of Direction Cosines Let           v = ai + bj + ck be a vector, then we define the direction cosines to be the following:                   a cos a  =                             ||v||                   b cos b  =                             ||v||                   c cos g  =                             ||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.   hence  We see that                                                       ||u|| ||v|| ||s||             u . v  =  ||u|| ||v|| cos q   =                          =  ||v|| ||s||                                                          ||u||                u . v   ||s||  =                                ||v|| We can calculate the projection of u onto v by the formula:                   u . v projvu =                v                    ||v||2   Work The work done by a constant force F along PQ is given by W = F . PQ    Example Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7)?  Solution We have that the force vector is          F  =  ma  =  (10)(-9.8j)  =  -98j and the displacement vector is         v  =  (5 - 2) i  +  (7 - 3) j  =  3i + 4j The work is the dot product         W  =  F . v  =  (-98j) . (3i + 4j)           =  (0)(3) + (-98)(4)  =  -392 Notice the negative sign verifies that the work is done against gravity.  Hence, it takes 392 J of work to move the baby.   Torque Suppose you are skiing and have a terrible fall.  Your body spins around and you ski stays in place (do not try this at home).  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. 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:                                  We can compute this determinant as                 =  (bf - ce) i + (cd - af) j + (ae - bd) k Example Find the cross product u x v if         u  =  2i + j - 3k            v  =  4j + 5k   Solution We calculate                 =  17i - 10j + 8k          If you need more help see the lecture notes for Math 103 B on matrices. Exercises Find u x v when u = 3i + j - 2k,      v = i - k u = 2i - 4j - k,      v = 3i -  j + 2k Notice that since switching the order of two rows of a determinant changes the sign of the determinant, we have          u x v  =  -v x u 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  Note:   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 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         20 cos 30  i  +  20 sin 30  j  = 17.3 i  +  10 j and the force as         -40 cos 75 i - 40 sin 75 j = -10.3 i - 38.6 j hence, the torque is the magnitude of their cross product:                 =  -564 inch pounds 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) This can be found by computing the determinate of the three vectors:       Example Find the volume of the parallelepiped spanned by the vectors         u  =  <1,0,2>        v  =  <0,2,3>        w  =  <0,1,3> Solution We find