I'm reading Hestenes' book "New Foundations of Classical Mechanics" as an introduction to Geometric (Clifford) Algebra. Don't worry, no physics mentioned here :)

An exercise asks to solve a vector equation, $$\alpha \boldsymbol x + \boldsymbol a (\boldsymbol x \cdot \boldsymbol b)=\boldsymbol c$$

The solution is given in the back of the book, yet I have absolutely no idea how to tackle this problem. FYI, the solution is: $$\boldsymbol x = \frac{\boldsymbol c}{\alpha} - \frac{(\boldsymbol c \cdot \boldsymbol b)\boldsymbol a}{\alpha(\alpha+\boldsymbol a\cdot \boldsymbol b)}$$

I'm familiar (but not experienced) with "vector division" and most of the vector identities you can deduce from the geometric product definitions.

The next exercise is similar, and seems to imply there are tons of these types of exercises I should be able to solve. I have searched for a more introductory text to these types of things, but didn't find anything useful for the above.

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    $\begingroup$ It would help if you specified what kind of things all of your symbols are. I think you are indicating vectors with bold but it is somewhat hard (at least for me) to distinguish the bold from non-bold letters. I generally prefer to write $v_i$ for vectors and $c_i$ for scalars. Another option is to write $\vec{x}$. $\endgroup$ – Qiaochu Yuan Dec 8 '12 at 18:14
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    $\begingroup$ a,b, and c are vectors, $\alpha$ is a scalar. Juxtaposition is geometric product, dot is inner product. $\endgroup$ – rubenvb Dec 8 '12 at 18:18
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    $\begingroup$ I am not at all sure that this resource might be of some help to you, but there is no harm in trying: it is "Linear Algebra via Exterior Products" by S.Winitski: sites.google.com/site/winitzki/linalg $\endgroup$ – Giuseppe Negro Dec 8 '12 at 18:24
  • $\begingroup$ @GiuseppeNegro thanks for the additional reference. I'll be sure to check it out! $\endgroup$ – rubenvb Dec 9 '12 at 11:20

Take inner product with $\boldsymbol b$ on both sides and solve for $\boldsymbol x \cdot \boldsymbol b$. Replace $\boldsymbol x \cdot \boldsymbol b$ in the original equation by the solution just found. Solve for $\boldsymbol x$.

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    $\begingroup$ When you put it like that, it seems so simple... Thanks. $\endgroup$ – rubenvb Dec 9 '12 at 11:20

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