Simplify the Dot Product in terms of $a$ and $b$

Question

Where $a$ and $b$ are arbitrary vectors $(a+2b) \cdot (2a-b)$

$$a\cdot(2a-b)+2b\cdot(2a-b) = 2(a\cdot a)-a\cdot b+1(b\cdot a)-2(b\cdot b)$$ $$=2(a)-ab+4ab-2(b)^2$$ $$=2a^2-2b^2$$ $$=2(a^2-b^2)$$

Where did i go wrong in simplifying this?

Answer 1

If you denote dot product as $\cdot$, you get $$ (a+2b) \cdot (2a - b) = 2 a \cdot a + 4 b \cdot a - a \cdot b - 2 b \cdot b = 2 a \cdot a + 3 b \cdot a - 2 b \cdot b $$ and you can further simplify $a\cdot a = |a|$ if you like.

Your error is the 2nd step (3rd line), where you assumed $$2b\cdot(2a-b) = 1 b\cdot a - 2 b \cdot b,$$ the the correct coefficient of $b \cdot a$ is 4, not 1.

Answer 2

I marked the mistakes in red and corrected these

\begin{align} (a+2b) * (2a-b)& =a*(2a-b)+2b*(2a-b)\\ & = 2(a*a)-a*b+\color{red}{4}(b*a)-2(b*b)\\ & = 2(a)\color{red}{^2}\underbrace{-ab+4ab}_{\color{red}{3ab}}-2(b)^2\\ & = 2a^2-2b^2+\color{red}{3ab}\\ & = 2(a^2-b^2)+\color{red}{3ab} \end{align}

Answer 3

$$(\mathbf a+2\mathbf b)\cdot(2\mathbf a-\mathbf b)=2\mathbf a\mathbf a-\mathbf a\mathbf b+4\mathbf b\mathbf a-2\mathbf b\mathbf b=2(|\mathbf a|^2-|\mathbf b|^2)+3\mathbf a\mathbf b$$