How do you find the cross product of #<2,3,0># and #2<-1,2,4>#?

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Answer 1 · 2016-12-16T15:42:55

The answer is #=〈12,-8,7〉#

The cross product of 2 vecors, #〈a,b,c〉# and #〈d,e,f〉# is given by the determinant

# | (hati,hatj,hatk),(a,b,c), (d,e,f) |#

Here the 2 vectors are #〈2,3,0〉# and #〈-1,2,4〉#

The cross product is

# | (hati,hatj,hatk),(2,3,0), (-1,2,4) |#

#=hati(12-0)-hatj(8-0)+hatk(4+3)#

#=〈12,-8,7〉#

Verification by doing the dot products

#〈2,3,0〉.〈12,-8,7〉=24-24+0=0#

#〈-1,2,4〉.〈12,-8,7〉=-12-16+28=0#

Since the dot products are #=0#, the vector obtained is perpendicular to the original vectors