Verify if $\overrightarrow{w}$ is a linear combination of $\overrightarrow{u}$ and $\overrightarrow{v}$

Question

I am given the following question:

Let $\Vert \overrightarrow{u} \Vert = \Vert \overrightarrow{v} \Vert = \Vert \overrightarrow{w} \Vert = 1$ and $\overrightarrow{u} \cdot \overrightarrow{v} = \overrightarrow{v} \cdot \overrightarrow{w} = \overrightarrow{u} \cdot \overrightarrow{w} = \frac{1}{2}$. Verify if $\overrightarrow{w}$ is a linear combination of $\overrightarrow{u}$ and $\overrightarrow{v}$.

I am not sure how to address the question. What I have so far (and I'm not sure if it is helpful) is

$$ \left( \overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} \right) \cdot \left( \overrightarrow{u} + \overrightarrow{v} + \overrightarrow{w} \right) = \\ \Vert \overrightarrow{u} \Vert^2 + \Vert \overrightarrow{v} \Vert^2 + \Vert \overrightarrow{w} \Vert^2 + 2 (\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{u} \cdot \overrightarrow{w}) = \\ 3 + 2(\overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{v} \cdot \overrightarrow{w} + \overrightarrow{u} \cdot \overrightarrow{w}) = 6 $$

Does that prove anything?

Answer 1

If $$w=au+bv$$ Dot product by $u$: $$\frac{1}{2}=a+b\frac{1}{2}$$ Dot product by $v$: $$\frac{1}{2}=a\frac{1}{2}+b$$ Hence $$1=\frac{3}{2}(a+b)$$ $$a+b=\frac{2}{3}$$ But dot product by $w$ gives $$w\cdot w=1=a\times \frac{1}{2}+b\times \frac{1}{2}$$ $$a+b=2$$

Hence $w$ is not a linear combination of $u$ and $v$.

Answer 2

$$ \text{suppose} \quad \overrightarrow{w}=a\overrightarrow{u}+b\overrightarrow{v}. \\ \overrightarrow{u}\cdot \overrightarrow{w}=a+\frac{b}{2}=\frac{1}{2}. \\ \overrightarrow{v}\cdot \overrightarrow{w}=\frac{a}{2}+b=\frac{1}{2}. $$ Thus, we have $a=b=\frac{1}{3}$, $\|\overrightarrow{w}\|^2=\frac{1}{9}(\|\overrightarrow{u}\|^2+2\overrightarrow{u}\cdot \overrightarrow{v}+\|\overrightarrow{v}\|^2)=\frac{1}{3}\neq 1$. We cannot represent $\overrightarrow{w}$ as the linear combination of $\overrightarrow{u}$ and $\overrightarrow{v}$

Answer 3

Since $\|\overrightarrow{u}\| = 1, \|\overrightarrow{v}\| = 1, |u\cdot v| = \frac{1}{2}$, it follows that $\overrightarrow{u} \neq \overrightarrow{v}$. The vector $\overrightarrow{w}$ is a linear combination of $\overrightarrow{u}, \overrightarrow{v}$ if and only if the vector triple product $\overrightarrow{w}\times(\overrightarrow{u} \times \overrightarrow{v}) = 0$. But \begin{equation*} \overrightarrow{w}\times(\overrightarrow{u} \times \overrightarrow{v}) = (\overrightarrow{w}\cdot \overrightarrow{v}) \overrightarrow{u} -(\overrightarrow{w}\cdot\overrightarrow{u})\overrightarrow{v} = \frac{1}{2}(\overrightarrow{u}-\overrightarrow{v}) \neq 0 \end{equation*}