For which values of x, the vectors $v_1=(1,x^2,x)$ and $v=(1,1,1)$ span a plane in $\mathbb{R}^3$. For every such $x$, find an equation of the corresponding plane.
Am I correct in saying that the vectors span a plane if they are linearly dependent? In that case, the vectors are linearly dependent if $x = 1$ or $x= -1$. How can you find an equation of the corresponding plane if $x = 1$?
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No. If they are linearly dependent, then the span a line, not a plane. In fact, $v_1$ and $v$ span a plane if and only if they are linearly independent. And that occurs if and only if $x\neq1$. In that case, $v_1$ and $v$ span the plan$$\{(\alpha,\beta,\gamma)\in\mathbb R^3\mid-x\alpha-\beta+(x+1)\gamma=0\}.$$
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Here is a simple way to find an equation of the plane, when the vectors are linearly independent (which happens if $x\ne 1$ only, b.t.w.)
For the sake of clearness, I'll denote $v_1=(1,\xi^2,\xi)$. The vector $u=(x,y,z)$ is in the plane generated by $v_1$ and $v$ if and only if the vectors $v_1,v,u$ are linearly dependent, i.e. if their determinant $$\begin{vmatrix} x&1&1\\ y&\xi^2&1 \\ z&\xi&1 \end{vmatrix}=0.$$
A plane is a vector space of dimension $\mathbf{2}$. So, you would need the vectors to be linearly indepedent.
For $2$ vectors, it is easy to test linear independence. For if they are linearly dependent, then one of them is a scalar multiple of the other.
Note that your $v_1$ already has the first coordinate $1.$ So the only way they are linearly dependent is if all the coordinates are equal. This directly gives you that $x = 1.$
Thus, the two vectors span a plane in $\mathbb{R}^3$ if and only iff $x \in \mathbb{R} \setminus \{1\}$.
In this case, you can find a vector perpendicular to both $v_1$ and $v.$
One of them would be $v_1 \times v = (x^2 - x, x-1, 1 - x^2).$
The plane will be given as $(x^2 - x) X + (x - 1)Y + (1 - x^2)Z = 0.$ (Note that the origin is a point lying on the plane.)
Another way to represent the plane would be: $t v_1 + uv$ where $t, u \in \mathbb{R}$.