How to prove the distributive law ( i.e $A.(B+C) = (A.B) + (A.C)$) when three vectors $A , B , C$ are not in the same plane?
I was reading the chapter called Multiplication of vectors. I had a hard time to understand the distributive law of dot product when $A , B , C$ are not in the same plane.
Everywhere that law is proved supposing three vectors are co planer.
Can anyone please help me by giving a comprehensive proof.
Best Answer
It seems that you are using the definition $${\bf a}\cdot{\bf b}:=|{\bf a}|\,|{\bf b}|\cos\phi\ ,$$ where $\phi\in[0,\pi]$ denotes the (nonoriented) angle between ${\bf a}$ and ${\bf b}$, if both are nonzero. For a proof of the distributive law $${\bf a}\cdot({\bf x}+{\bf y})={\bf a}\cdot{\bf x}+{\bf a}\cdot{\bf y}\tag{1}$$ we may assume that $|{\bf a}|=1$. The proof rests on the following idea: Attach ${\bf a}$ at the origin ${\bf 0}$, and consider this vector as fixed. Let $g$ be the line through ${\bf 0}$ spanned by ${\bf a}$, and denote by $\pi_g:\>{\mathbb R}^3\to g$ the orthogonal projection of "everything" in space to $g$. You should convince yourself that this map, defined by a geometrical construction, is linear. Furthermore you can verify that due to $|{\bf a}|=1$ for any vector ${\bf x}$ attached at the origin we have $\pi_g({\bf x})=\lambda{\bf a}$ with $$\lambda=\cos\phi\,|{\bf x|}=|{\bf a|}\,|{\bf x|}\cos\phi={\bf a}\cdot{\bf x}\ ,$$ hence $$\pi_g({\bf x})=({\bf a}\cdot{\bf x})\>{\bf a}\qquad\forall {\bf x}\in{\mathbb R}^3\ .$$ The linearity of $\pi_g$, i.e., $\pi_g({\bf x}+{\bf y})=\pi_g({\bf x})+\pi_g({\bf y})$, then implies $(1)$.