What is the cross product of $<3 ,1 ,-6 >$ and $<7 ,3 ,2 >$ ?

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Answer 1 · 2016-12-26T13:43:07

The cross product is $=〈20,-48,2〉$

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 $〈3,1,-6〉$ and $〈7,3,2〉$

And the cross product is

$| (hati,hatj,hatk), (3,1,-6), (7,3,2) |$

$=hati| (1,-6), (3,2) | - hatj| (3,-6), (7,2) |+hatk | (3,1), (7,3) |$

$=hati(2+18)-hati(6+42)+hatk(9-7)$

$=〈20,-48,2〉$

Verification, by doing the dot product

$〈20,-48,2〉.〈3,1,-6〉=60-48-12=0$

$〈20,-48,2〉.〈7,3,2〉=140-144+4=0$

Therefore, the vector is perpendicular to the other 2 vectors

What is the cross product of <3 ,1 ,-6 ><3 ,1 ,-6 > and <7 ,3 ,2 ><7 ,3 ,2 >?