# 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` .

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

`A=[[1,3,x],[4,1,y],[2,-1,z]]`

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` .