Show that the "calculation rule" is wrong for the given scalar product

Question

Show that the "calculation rule" is wrong for the scalar product $\left\langle\vec{u},\vec{v}\right\rangle\cdot\vec{w}=\vec{u}\cdot\left\langle\vec{v},\vec{w}\right\rangle$

I have problem showing that because I actually get a true result : /

I did:

Let $\vec{u}= \begin{pmatrix} u_1\\ u_2\\ u_3 \end{pmatrix}$, let $\vec{v}= \begin{pmatrix} v_1\\ v_2\\ v_3 \end{pmatrix}$, $\vec{w} = \begin{pmatrix} w_1\\ w_2\\ w_3 \end{pmatrix}$

Then $$\left\langle\vec{u},\vec{v}\right\rangle\cdot\vec{w}=\vec{u}\cdot\left\langle\vec{v},\vec{w}\right\rangle \Leftrightarrow (u_1v_1+u_2v_2+u_3v_3) \begin{pmatrix} w_1\\ w_2\\ w_3 \end{pmatrix} = \begin{pmatrix} u_1\\ u_2\\ u_3 \end{pmatrix}(v_1w_1+v_2w_2+v_3w_3) \Leftrightarrow$$ $$\Leftrightarrow\begin{pmatrix} u_1v_1w_1\\ u_2v_2w_2\\ u_3v_3w_3 \end{pmatrix} = \begin{pmatrix} u_1v_1w_1\\ u_2v_2w_2\\ u_3v_3w_3 \end{pmatrix}$$

I don't see why it must be wrong? :s

Answer 1

Your last equivalence is false. We don’t have :

$$(u_1v_1+u_2v_2+u_3v_3) \cdot w_1 = u_1v_1w_1$$

Because : $(u_1v_1+ u_2v_2+u_3v_3)$ is a scalar ( a $1 \times 1$ vector).

Now can you continue your proof ?

Answer 2

You're making this much harder than necessary. To disprove a rule, you need merely show one counterexample. Here is one:

$$\vec{u}= \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \quad \vec{v}= \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \quad \vec{w} = \begin{pmatrix} 0\\1\\0 \end{pmatrix}.$$