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

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Answer 1 · 2016-12-16T15:42:55

The answer is $= ⟨ 12, - 8, 7 ⟩$ $=〈12,-8,7〉$

The cross product of 2 vecors, $⟨ a, b, c ⟩$ $〈a,b,c〉$ and $⟨ d, e, f ⟩$ $〈d,e,f〉$ 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

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