Let `vecv = <1,4,2>` and `vec w = <3,1,-1>` . Find conditions on `x, y, z` that ensure that `<x, y, z>`  is in the span of `vecv` and `vecw` . 

1 Answer

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

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Anything in the span of `vecv` and `vecw` is of the form

`a[[1],[4],[2]]+b[[3],[1],[-1]]=[[a+3b],[4a+b],[2a-b]],` so `[[x],[y],[z]]` will be in this span if and only if

`x=a+3b,`   `y=4a+b,`   and `z=2a-b` for some numbers `a` and `b.`

There's another way to state this condition. Recall that the determinant of the matrix


is zero if and only if the columns are linearly dependent. In other words, the determinant is zero if and only if

`r[[1],[4],[2]]+s[[3],[1],[-1]]+t[[x],[y],[z]]=vec 0,`

where at least one of `r,s,t` does not equal zero. But note that `t` can't be zero, since then a nontrivial combination of the vectors `vecv` and `vecw` would give the zero vector. Since `vecv` and `vecw` are independent, this can't happen. Thus `t!=0,` and we can solve for `[[x],[y],[z]]` in terms of `vecv` and `vecw.` But this just means `[[x],[y],[z]]` is in the span of `vecv` and `vecw` if and only if the determinant of matrix `A` is zero.

By cofactor expansion down the last column, we see that this determinant is `-6x+7y-11z,` and so we get the equivalent condition that `[[x],[y],[z]]` is in the span if and only if `-6x+7y-11z=0.`

This can also be viewed a third way, by finding the equation of the plane formed by `vecv` and `vecw` .