prove that the $\{ \vec{v_1} , \vec{v_2} , \operatorname{proj}_n\vec{x}\}$ is linearly independent for all $\vec{x} \in R^3, \vec{x}\notin P$.

Question

Assume $P = \operatorname{span}\{ \vec{v_1} , \vec{v_2}\}$ be a plane in $R^3$ with a normal vector $\vec{n}$. I want to prove that the $\{ \vec{v_1} , \vec{v_2} , \operatorname{proj}_n\vec{x}\}$ is linearly independent for all $\vec{x} \in R^3, \vec{x}\notin P$.

My proof:

We know that $\{ \vec{v_1} , \vec{v_2}\}$ is L.I. since it is a plane in $R^3$. We need to show that $c_3 =0$: $$c_1\vec{v_1} + c_2\vec{v_2} + c_3\operatorname{proj}_n\vec{x} = 0$$

We see that $\operatorname{proj}_n\vec{x}$ is a scaler multuple of the normal for the plane $P$, hence since $\operatorname{proj}_n\vec{x}$ will be orthogonal to both $\{ \vec{v_1} , \vec{v_2}\}$ you can never write $\operatorname{proj}_n\vec{x}$ in terms of $\vec{v_1}$ and $\vec{v_2}$. Hence $c_3 =0$.

Is this a valid proof?

Answer 1

It depends who is looking at your proof. You proof is perfectly good for most people, but maybe not for a university assignment!

This is a more concise statement of your proof as it is:

Any normal vector $\vec{n}$ of the plane is linearly independent to $\vec{v_1}$ and $\vec{v_2}$. Note that for any $\vec{x} \in \mathbb{R}^3$ not in $P$, its projection onto any normal vector $\vec{n}$ of $P$ is a non-zero scalar multiple of $\vec{n}$, hence also a normal vector. Hence linearly independent with $\vec{v_1}$ and $\vec{v_2}$

This may not get you full marks on a school assignment though, because you haven't proved the statement:

Any normal vector $\vec{n}$ of a plane $P = Span \{ \vec{v_1} , \vec{v_2} \} \subseteq \mathbb{R}^3 $ is linearly independent to $\vec{v_1}$ and $\vec{v_2}$

To prove this, you need that $$c_1\vec{v_1} + c_2\vec{v_2} + c_3 \vec{n} = 0 \iff c_1 = c_2 = c_3 = 0 $$ Well if $c_3 = 0$, then $c_1 = c_2 = 0$ by the linear independence of $\vec{v_1}$ and $\vec{v_2}$. So assume that $c_3 \neq 0$. Then $$ c_1\vec{v_1} + c_2\vec{v_2} + c_3 \vec{n} = 0 $$ $$ \implies -\frac{c_1}{c_3} \vec{v_1} + -\frac{c_2}{c_3}\vec{v_2} = \vec{n} $$ $$ \implies \vec{n} \in Span \{ \vec{v_1} , \vec{v_2} \} = P$$ which is a contradiction

Answer 2

It is more or less correct, but:

• The set $\left\{\vec{v_1},\vec{v_2}\right\}$ is not a plane. It spans a plane.
• You cannot just say, without justification, that something cannot be done. Yes, $\operatorname{proj}_{\vec n}\vec x$ cannot be written as a linear combination of $\vec{v_1}$ and $\vec{v_2}$ when $x\notin P$, but you have to say why. If $\vec x\notin P$, then $\operatorname{proj}_{\vec n}\vec x$ is a non-zero scalar multiple of $\vec n$ and therefore it follows from$$c_1\vec{v_1}+c_2\vec{v_2}+c_3\operatorname{proj}_{\vec n}\vec x=0$$that$$\left\langle c_1\vec{v_1}+c_2\vec{v_2}+c_3\operatorname{proj}_{\vec n}\vec x,\operatorname{proj}_{\vec n}\vec x\right\rangle=0,$$but $\left\langle c_1\vec{v_1}+c_2\vec{v_2}+c_3\operatorname{proj}_{\vec n}\vec x,\operatorname{proj}_{\vec n}\vec x\right\rangle=c_3\left\lVert\operatorname{proj}_{\vec n}\vec x\right\rVert^2$ and therefore, since $\left\lVert\operatorname{proj}_{\vec n}\vec x\right\rVert^2\neq0$, $c_3=0$.