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 Post subject: Vector equationPosted: Tue Nov 10, 2015 9:09 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let $\mathbf{a, b, c} \in \mathbb{R}^3$ be three vectors. Solve the equation:

$$\mathbf{x+(x\cdot a)b = c}$$

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 Post subject: Re: Vector equationPosted: Fri Jan 15, 2016 5:26 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Tolaso J Kos wrote:
Let $\mathbf{a, b, c} \in \mathbb{R}^3$ be three vectors. Solve the equation:

$$\mathbf{x+(x\cdot a)b = c}$$

Here is a solution:

Successively we have:

\begin{align*} \mathbf{x+ (x\cdot a)b=c} &\Rightarrow \mathbf{x \cdot a+ (x\cdot a)\cdot (a\cdot b)=a\cdot c} \\ &\Rightarrow \mathbf{x\cdot a= \frac{a\cdot c}{1+a\cdot b}} \\ &\Rightarrow \mathbf{x=c - \frac{a\cdot c}{1+a\cdot b}b} \end{align*}

The last solution verifies the initial if an only if $1+ \mathbf{a\cdot b} \neq 0$. We are investigating further.

Investigation

If $\mathbf{1+a\cdot b}=0$ then $\mathbf{x}=m\mathbf{b}+n \mathbf{c}$ so the initial equation is transfmored into:

$$m\mathbf{b}+n \mathbf{c}+ \left ( -m +n \cdot \mathbf{a\cdot c} \right )\cdot \mathbf{b}=\mathbf{c}$$

or in a more simplified form: $\displaystyle n \mathbf{c}+ n \mathbf{(a\cdot c)\cdot b}=\mathbf{c}$.

• If $\mathbf{a \cdot c}=0$ then $n=1$ hence $\mathbf{x}=m \mathbf{b} +\mathbf{c} , \;\; m \in \mathbb{R}$.
• If $\mathbf{a \cdot c}\neq 0$ and $\mathbf{b, c}$ not collinear then the initial equation is obviously impossible .
• If $\mathbf{a\cdot c} \neq 0$ and $\mathbf{b, c}$ are collinear then the equation is transformed to: $$k \mathbf{b}+ \left ( k \mathbf{b \cdot a} \right )\cdot b = \ell \mathbf{b}\Rightarrow \ell \mathbf{b}=0 \Rightarrow \mathbf{b}=0 \; {\rm or} \; \mathbf{c}=0$$

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