Question

Problem 6: Problem 1.86 in Young &Freedman Later in our study of physics we will encounuanties represented by (AxB) c. (a) Prove that for any three vectors A, B, and C, the equality A . (BX C)-ИХ B) . C holds (b) Calculate (AX B) . C for vector A with magnitude A-5.00 and angle θ,-26.00 (measured from the +x-axis toward the +y-axis), vector B with B = 4.00 and θ,- 63.0% and vector C with magnitude 6.00 and in the +z-direction. Vectors A and B are in the xy-plane

Answer 1

for this

i.i = j.j = k.k = 1

ixi = jxj = kxk = 0

For this I middot i = j middot j = k middot k = 1 I times i = j times j = k times k = 0 Let A^vector = A x i^^+ A y j^^+ A z k^^B^vector = B x i^^+ B y j^^+ B z k^^& C^vector = C x i^^+ C y j^^+ C z k^^Now B^vector times C^vector = Vertical-bar i^^B_x C_x j B_y C_y k^^B_z C_z Vertical-bar = i(B_y C_z - B_z C_y) - j^^(B_x C_z - B_z C_x) + k^^(B_x C_y - B_y C_x) \r\n consider LHS of the identity LHS = A^vector middot (B^vector times C^vector) = (A_x i^^+ A_y j^^+ A_z k^^) middot [i (B_y C_z - B_z C_y) - j^^(B_z C_z - B - (x) + k^^(B_x C_y - B_y C_x)] = A_x (B_y C Vertical-bar _z - B_z C_y) - A_y (B_x C?z - B_z C_x) + A_z (B_x (y - B_y(x) = A_x B_y C_z - A_x B_z

b)

the direction of cross product is always in the direction perpendicular to the plane containing the vectors

in part b A and B are in xy plane so A x B is in the z direction

C is also in z direction i.e. both are in the direction of unit vector k

hence (A x B) . C = (12.036 k).(6k) = 72.22