Hints only
Let $P = \text{Span } \{v_1, v_2\}$ be a plane in $\mathbb{R}^3$ with normal vector $n$. Show that $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. Hints only
Equation for $P$: $P = c_1v_1 + c_2v_2$. For real $c_1, c_2$. We have by definition, $n = v_1 \times v_2$.
To make sure $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. we must have
- Span $\{ v_1, v_2, n\}$ = $\mathbb{R}^3$
- $\{ v_1, v_2, n\}$ Linearly independent.
I am having major trouble showing (1).
Can I get a SMALL hint?
I am not allowed to use dimension.
Showing set is LI
By definition of a plane, $\{v_1, v_2 \}$ is linearly independent thus $\overrightarrow{0} \not \in \{v_1, v_2 \}$
So $c_1v_1 + c_2v_2 + 0 \overrightarrow{n} = 0$ Thus $\{v_1, v_2, n \}$ is LI.
The span part is confusing
Best Answer
You want to show that $\{ v_1, v_2, n\}$ is a basis, meaning it is a linearly-independent set generating all of $\mathbb{R}^3$.
Linear independency means that you need to show that the only way to get the zero vector is by the null linear combination.
Now what is left is to show the generating part meaning: span$\{ v_1, v_2, n\}=\mathbb{R}^3$.
Obviously, span$\{ v_1, v_2, n\}\subseteq \mathbb{R}^3$ by definition of span and definition of a vector space.
All you need to show is that $\mathbb{R}^3\subseteq$ span$\{ v_1, v_2, n\}$. That would mean that an arbitrary vector in $\mathbb{R}^3$ can be written as a linear combination of $v_1$, $v_2$ and $n$.
This is the long way to prove it (recommended to understand). Once you are familiar with this, the fast way is to use the fact that a span is a vector subspace and a lemma that states if a vector subspace $W\subseteq V$ has the same dimension as $V$ then $W=V$.