(the scalar projection of F onto d): Thus, the work done by the force to displace the particle from Suppose this is not the case. Here, So, be careful with notation and make sure you are finding the correct projection. cross in Here it is, Note that we also need to be very careful with notation here. is |b|cos(theta) (where theta is the angle between a and orthogonal vectors is zero. Given the two vectors \(\vec a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \) and \(\vec b = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \) the dot product is. direction (hence the notation comp). Not much to do with these other than use the formula. Here are a couple of sketches illustrating the projection. The proofs of these properties are mostly “computational” proofs and so we’re only going to do a couple of them and leave the rest to you to prove. This in turn however means that we must have \({v_i} = 0\) and so we must have had \(\vec v = \vec 0\). Conversely, the only way the dot product vector be F=<2,3,4> and the displacement vector be The symbol for dot product is represented by a heavy dot (.) b=<-2,4,-1> is. of the vector projection of b onto a. An important use of the dot product is to test whether or not two vectors distance moved (the magnitude of the displacement vector) and the magnitude b=

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