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Imagine two vectors a and b that point in the same direction In this situation, the angle qab between the vectors is equal to 0 If the magnitude of a is ra and the magnitude of b is rb, then the dot product is a b = rarb cos qab = rarb cos 0 = rarb 1 = rarb Now think of two vectors c and d that point in opposite directions The angle qcd between the vectors is equal to p If the magnitude of c is rc and the magnitude of d is rd, then the dot product is c d = rcrd cos qcd = rcrd cos p = rcrd ( 1) = rcrd
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144 Vectors in Cartesian Three-Space
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The cross product a b of two vectors a and b in three-dimensional space can be found according to the same rules we learned for finding a cross product in polar two-space We get a vector perpendicular to the plane containing a and b, and whose magnitude ra b is given by ra b = rarb sin qab where ra is the magnitude of a, rb is the magnitude of b, and qab is the angle between a and b, expressed in the rotational sense going from a to b When we want to figure out a cross product, it s always best to keep the angle between the vectors nonnegative, but not larger than p That is, we should restrict the angle to the following range: 0 qab p If we look at vectors a and b from some vantage point far away from the plane containing them, and if qab turns through a half circle or less counterclockwise as we go from a to b, then a b points toward us If qab turns through a half circle or less clockwise as we go from a to b, then a b points away from us In any case, the cross product vector is precisely perpendicular to both the original vectors
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An example Consider two vectors a and b in three-space Imagine that they both have magnitude 2, but their directions differ by p /6 We can plug the numbers into the formula for the magnitude of the cross product of two vectors, and calculate as follows:
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ra b = rarb sin qab = 2 2 sin (p /6) = 4 1/2 = 2 If the p /6 angular rotation from a to b goes counterclockwise as we observe it, then a b points toward us If the p /6 angular rotation from a to b goes clockwise as we see it, then a b points away from us
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Another example Now think about two vectors c and d, represented by ordered triples as
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c = (1,1,1) and d = ( 2, 2, 2) Let s find the cross product c d From the information we ve been given, we can see immediately that d = -2c That means the magnitude of d is twice the magnitude of c, and the two vectors point in opposite directions We can calculate the magnitude rc of vector c as rc = (12 + 12 + 12)1/2 = (1 + 1 + 1)1/2 = 31/2
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and the magnitude rd of vector d as rd = [( 2)2 + ( 2)2 + ( 2)2]1/2 = (4 + 4 + 4)1/2 = 121/2 When two vectors point in opposite directions, the angle between them is p, whether we go clockwise or counterclockwise We now have all the information we need to figure out the magnitude rc d of the cross product c d using the formula rc d = rcrd sin qcd = 31/2 121/2 sin p = 31/2 121/2 0 = 0 The cross product c d is the zero vector, because its magnitude is 0 Although we don t yet have a formula for figuring out cross products directly from ordered triples in xyz space, we can infer from this result that (1,1,1) ( 2, 2, 2) = (0,0,0) where the bold times sign ( ) denotes the cross product, not ordinary multiplication