calculus properties of cross products?

NetherCraft 0

Find the vector, not with determinants, but by using properties of cross products. (j – k) x (k – i)

2 Answers

  • use the distributive properties of the cross product:

    (j – k) x (k – i) = (j-k) x k – (j-k) x i =

    j x k – k x k – ( j x i – k x i ) = j x k – k x k – j x i + k x i

    Now k x k = 0

    The other 3 you can do by visualizing it geometrically: draw them out and use the right hand rule for the direction. The length of the vector that results in each of these three must = 1 since the norm of a x b is |a||b|sinΘ, and in all these 3 cases we have |i|=|j|=|k| = 1 and Θ= pi/2 so sin Θ = 1. :

    j x k = i

    j x i = – k

    k x i = j

    So we get i – (-k) + j = i + j + k

  • Cross Products Property

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