What is the cross product of $<4 , 5 ,-9 >$ and $<4, 8 ,-2 >$ ?

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Answer 1 · 2016-12-27T15:18:46

The answer is $=〈62,-28,12〉$

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

is given by the determinant

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

and $| (a,b), (c,d) |=ad-bc$

Here, the 2 vectors are $〈4,5,-9〉$ and $〈4,8,-2〉$

And the cross product is

$| (hati,hatj,hatk), (4,5,-9), (4,8,-2) |$

$=hati| (5,-9), (8,-2) | - hatj| (4,-9), (4,-2) |+hatk | (4,5), (4,8) |$

$=hati(-10+72)-hati(-8+36)+hatk(32-20)$

$=〈62,-28,12〉$

Verification, by doing the dot product

$〈62,-28,12〉.〈4,5,-9〉=248-140-108=0$

$〈62,-28,12〉.〈4,8,-2〉=248-224-24=0$

Therefore, the vector is perpendicular to the other 2 vectors

What is the cross product of <4 , 5 ,-9 ><4 , 5 ,-9 > and <4, 8 ,-2 ><4, 8 ,-2 >?