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I recently read a formula for dot product of two vectors in my course book. Let there be two vectors $\vec{a}$ and $\vec{b}$ and $c \in \mathbb{R}$.

It stated that $$c\vec{a} \cdot c\vec{b} = c(\vec{a} \cdot\vec{b})$$

I'm really confused why the real number isn't squared in the RHS. According to my understanding of multiplication, the RHS should look like $c^2(\vec{a} \cdot\vec{b})$. Am I making some big blunder here or maybe there is a printing/typing mistake in the book i'm referring to?

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    $\begingroup$ You are correct, and there is a printing/typing mistake in the book. $\endgroup$ – Rory Daulton Dec 31 '15 at 13:46
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The dot product is bilinear, meaning that for all $x,y,z \in \mathbb{R}^n$ and $\lambda \in \mathbb{R}$ we have $$ x \cdot (y+z) = x \cdot y + x \cdot z, \\ (x+y) \cdot z = x \cdot z + y \cdot z, \\ (\lambda x) \cdot y = \lambda (x \cdot y) = x \cdot (\lambda y). $$ From the last equality it follows that for $a,b \in \mathbb{R}^n$ and $c \in \mathbb{R}$ $$ (ca) \cdot (cb) = c (a \cdot (cb)) = c^2 (a \cdot b), $$ so you are right.

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  • $\begingroup$ The book is a very reputed one used by more than half the schools here and in my last 5 years with it i had never encountered a typo so i had to confirm it here before marking it as a typo and your steps loook really sound to me so i would take is at a typo. Thank you very much. $\endgroup$ – user301926 Dec 31 '15 at 14:48

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