Let `vecv = <1,4,2>` and `vec w = <3,1,-1>` . Compute `vecv xx vecw` and use it to find an equation for the plane determined by `vecv` and `vecw.`

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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The cross product

`<a_1,a_2,a_3>xx<b_1,b_2,b_3> `

`=<a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1>.` In this case, we have `a_1=1,a_2=4,a_3=2,b_1=3,b_2=1,b_3=-1` and when we plug these into the above formula, we see that

`<1,4,2>xx<3,1,-1> = <-6,7,-11>.`

The key fact here about the cross product is that it's normal (perpendicular) to the plane formed by the vectors `vecv` and `vecw.` That is equivalent to the dot product of this normal vector and any vector lying in the plane being zero, in other words

`<-6,7,-11>*<x,y,z> =-6x+7y-11z=0,`

so the equation of the plane is `-6x+7y-11z=0.`

Compare this result to the answer to this question .