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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 $\mathbb{R}^3$

It must be that $\{v_1, v_2\}$ is linearly independent (LI) by def of a plane, thus $c_1v_1 + c_2v_2 = \overrightarrow{0}$ for $c_1 = c_2 = 0$.

It must follow that the coefficient of $n$ is 0?

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    @Mathematician42, shucks! I forgot I had already asked this question. So sorry about this. What can I do now? – 2017-02-05
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    No idea, you could have read the answers to your other question, in that case you would know the answer to this and there is no need to ask it again. – 2017-02-05

2 Answers 2

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$v_1$ and $v_2$ must be independent since the span a plane. Consider \begin{align*} c_1v_1+c_2v_2+c_3n & = \vec{0}\\ c_1(v_1 \cdot n)+c_2(v_2 \cdot n)+c_3(n \cdot n) & = (\vec{0} \cdot n)\\ \end{align*} By orthogonality we get $c_3=0$. So we are left with $$c_1v_1+c_2v_2 = \vec{0}.$$ Now use the independence of $v_1$ and $v_2$.

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    Why cant I just say since $c_1v_1 + c_2v_2 = 0$ and $c_1 = c_2 = 0$, if we have $c_1v_1 + c_2v_2 + c_3v_3 = 0$, must have $c_3 = 0$? – 2017-02-05
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    I mean $0v_1 + 0v_2 = 0$ thus only possibility of addition is $0v_3$ – 2017-02-05
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    @Amad27 two vectors could be linearly independent but when you have a third vector then the new set may no longer be independent. For example take $v_1=(1,0,0), v_2=(0,1,0)$ and $n=(1,1,0)$. Then $c_1v_1+c_2v_2=0$ will be satisfied only with $c_1=c_2=0$ but when we create a linear combination of the three vectors, then $1v_1+1v_2-1n=\vec{0}$. – 2017-02-05
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Hint: Assume $$c_1v_1+c_2v_2+c_3n=0,$$ now take the inner/scalar product with $n$.

Conclude