It can be shown that the cross product of two vectors, A = A_x i + A_y j + A_z k

Question

It can be shown that the cross product of two vectors, A = A_x i + A_y j + A_z k, and B = B_x i + B_y j + B_z k is as follows: A x B = (A_yB_z - A_zB_y) T + (A_ZB_X - A_XB_Z) j + (A_xB_y - A_yB_x) k The cross product can then be written as follows, where we use the rules for evaluating a determinant. (Note, however, that this is not actually a determinant, but a memory aid.) Assume that A = 8.0 i - 5.2 j and B = -8.5 i + 7.0 j + 2.0 k. Use the above results to determine the vector product A x B. Use the above results to determine the angle between A and B.

Explanation / Answer

a) A cross B = i*( (-5.2)*2 - 7*0) - j*( 8*2 - (-8.5)*0) + k*(8*7 - (-8.5)*(-5.2))

= -10.4i - 16j + 11.8k

b) we know,

|A| = sqrt(8^2 + 5.2^2 + 0^2) = 9.54

|B| = sqrt(8.5^2 + 7^2 + 2^2) = 11.19

|A cross B| = sqrt(10.4^2 + 16^2 + 11.8^2) = 22.4

|A cross B| = |A|*|B|*sin(theta)

22.4 = 9.54*11.19*sin(theta)

sin(theta) = 0.2098

theta = sin^-1(0.2098)

= 12.1 degrees

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