The Scalar ProductThe vectors A and B are given by A = 31 + lj and B = -1î + 1j.(a) Determine the scalar product A- B.SOLUTIONConceptualize There is no physical system to imagine here. Rather, it is purely a mathematical exercise involving two (vectors vCategorize Because we have a definition for the scalar product, we categorize this example as (a substitution v problem.Substitute the specific vector expressions for A and B:A-B - (31 + 1j) - (-1i + 1j)- [3i - (-1)1 - [31 - 15 - [13 - (-1)1 + [19 - 15= |-2The same result is obtained when we use the equation A- B = A B + A B +AB, directly, where A, = 3, A, = 1, B, = -1 and B = 1.(b) Find the angle e between A and B.SOLUTIONEvaluate the magnitudes of A and B using the Pythagorean theorem:A = VA? + A 2 = AB = VB 2 +B 2 =0Use the equation A - B = AB cos(e) and the result from part (a) to find the angle (in degrees).A-Bcos(e) == 116.8ABEXERCISEAs a particle moves from the origin to (7i + 5j) m, it is acted upon by a force given by (7î + 7j) N. Calculate the work done by this force on the particle as it moves through the given displacement. (Give your answer in J.)HintW = 14Use the expression for the work done by a force acting on an object as it is displaced to calculate the work done. J

(b) The magnitude of vectors A and B is, A=Ax2+Ay2=32+12=10=3.162B=Bx2+By2=-12+12=2=1.414Exercise The work done is equal to, W=F·r=7 i+7 j N·7i+5j m=49+35=84 J

Answered: (b) Find the angle e between A and B.…

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