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

1 Answer
Dec 27, 2016

Answer:

The answer is #=〈62,-28,12〉#

Explanation:

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) | #

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

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