Proving $\left\{\vec{a},\vec{b}, \operatorname{proj}_\vec{n}(\vec{x})\right\}$ is linearly independent

Question

$P$ is. plane in $\Bbb{R^3}$ with equation $\operatorname{span} \left\{ \vec{a}, \vec{b} \right\}$ how would you prove that $\left\{\vec{a},\vec{b}, \operatorname{proj}_\vec{n} (\vec{x}) \right\}$ is linearly independent for all $\vec{x} \in \Bbb{R^3}, \vec{x} \not\in P$ where $\vec{n}$ is the normal vector of the plane?

I tried doing contradiction and contrapositive but have not been able to get anywhere with that, any help would be appreciated.

Is it possible to just say since ${proj}_\vec{n}(\vec{x})$ is not $\vec{0}$ and that it is just a scalar multiple of $\vec{n}$ then $\vec{a}$ and $\vec{b}$ are orthogonal to ${proj}_\vec{n}(\vec{x})$ so the set must be linearly indpendent

Answer 1

Hint By construction $\operatorname{proj}_{\bf n} {\bf x}$ is a multiple of $\bf n$, and since ${\bf x} \not\in P$, it is a nonzero multiple. Therefore the problem is equivalent to showing that $({\bf a}, {\bf b}, {\bf n})$ is linearly independent. Now, since ${\bf a}, {\bf b}$ span $P$, they are linearly independent and ${\bf n}$ is some nonzero multiple of ${\bf a} \times {\bf b}$.

Alternatively, use the usual definition of linear independent: Suppose there are constants $\alpha, \beta, \gamma$ not all zero such that $$\alpha {\bf a} + \beta {\bf b} + \gamma \operatorname{proj}_{\bf n} {\bf x} = {\bf 0} .$$ Now form the dot product of both sides with appropriate vector(s).

Answer 2

$\textbf{Proof}$. Assume that $\vec{x}$ is a nonzero vector such that $\vec{x}\not= \: u\vec{a} + v\vec{b}$ for any $(u,v)\in \mathbb{R}^2$. Since $$ \text{proj}_{\vec{n}}\vec{x} = \frac{\vec{n}\cdot \vec{x}}{\vec{n}\cdot \vec{n}}\:\vec{n} = c\: \vec{n}, $$ where $c\not=0$, the vector $\text{proj}_{\vec{n}}\vec{x}$ is a scalar multiple of $\vec{n}$; that is, $\text{proj}_{\vec{n}}\vec{x}\: ||\: \vec{n}$.

Since $\vec{a}\times \vec{b}=d\: \vec{n}$, where $d\not=0$, $$ \text{proj}_{\vec{n}}\vec{x} = \frac{c}{d}\: \vec{a}\times \vec{b} = c' \: \vec{a}\times \vec{b}, \hspace{4mm} \mbox{ where } c' = \frac{c}{d} \not=0. $$ Now, consider $$ A\vec{a} + B\vec{b} + C (\vec{a}\times \vec{b}) = \vec{0}. $$ Then dotting the previous vector equation by the vector $\vec{a}\times \vec{b}$, we obtain $$ \begin{align*} \left( A\vec{a} + B\vec{b} + C (\vec{a}\times \vec{b})\right) \cdot \vec{a}\times \vec{b} &= A\vec{a} \cdot (\vec{a}\times \vec{b}) + B\vec{b} \cdot (\vec{a}\times \vec{b}) + C ||\vec{a}\times \vec{b} ||^2 \\ &=0 + 0 + C ||\vec{a}\times \vec{b} ||^2 \\ &=0. \end{align*} $$ Since $||\vec{a}\times \vec{b} ||\not=0$, $C=0$.

So $A\vec{a} + B\vec{b}=\vec{0}$. Since the two vectors $\vec{a}, \vec{b}$ span the ($2$-dimensional) plane $P\subseteq \mathbb{R}^3$, $\vec{a}$ and $\vec{b}$ are linearly independent. So $A=B=0$.

Thus $\{ \vec{a},\vec{b}, \vec{a}\times \vec{b}\}$ is linearly independent implies $\{ \vec{a},\vec{b}, \text{proj}_{\vec{n}}\vec{x}\}$ is linearly independent. $\hspace{10mm}\square$