# find an angle in $\mathbb{R^3}$ between vectors defined by dot and cross products

First all, Merry Christmas to everyone!

I am stuck and need help!

Three vectors in $\mathbb R^3$, $\vec a$, $\vec b$ and $\vec c$, where $\vec b$ is a unit vector, and $\vec c$ is defined as

$\vec c$ = ($\vec a$ . $\vec b$) $\vec b$ + [$\vec a$ - ($\vec a$ . $\vec b$)$\vec b$] $cos(\alpha)$ + [$\vec a$ $\times$ $\vec b$] $sin(\alpha)$

What do the square brackets mean? how can the sum of a dot product and a cross product be a vector?

I want to find out an angle between $\vec c$ and $\vec b$. But I do not understand the equation of $\vec b$ and can't work out $\vec b$.

Thank you very much!

• how can the sum of a dot product and a cross product You are misreading the expression. All those are vectors on the RHS: $\vec c$ = ($\vec a$ . $\vec b$) $\color{red}{\vec b}$ + [$\color{red}{\vec a}$ - ($\vec a$ . $\vec b$)$\color{red}{\vec b}$] $cos(\alpha)$ + [$\color{red}{\vec a \times \vec b}$] $sin(\alpha)$. – dxiv Dec 28 '17 at 18:30
• You made me think the question had something to do with Christmas. – Yanko Dec 28 '17 at 18:31
• Thank you for your reply, dxiv. Can you help me understand this question? – Newbee Dec 28 '17 at 18:37
• Merry Christmas yanko – Newbee Dec 28 '17 at 18:37
• $$b\cdot c=(a\cdot b)(b\cdot b)+(a\cdot b)(1-b\cdot b)\cos\alpha$$ $$=a\cdot b$$ – velut luna Dec 28 '17 at 18:40

Hint: using that $\vec b \cdot \vec b = |\vec b|^2=1$ and $(\vec a \times \vec b) \cdot \vec b = 0\,$:
$$\require{cancel} \vec c \cdot \vec b = (\vec a \cdot \vec b) \,\vec b \cdot \vec b + (\cancel{\vec a \cdot \vec b} - \cancel{(\vec a \cdot \vec b) \,\vec b \cdot \vec b}) \cos(\alpha) + \xcancel{(\vec a \times \vec b) \cdot \vec b \sin(\alpha)} = \vec a \cdot \vec b$$