Let A and B be arbitrary vectors.
simplify
- $(a+2b)\cdot(2a-b)$
I did
$(2a\cdot a)-(a\cdot b)+(2a\cdot b)-2(b\cdot b)$
$2[a^2]+ab-2[b^2]$ would this be correct
2.$(a+2b)\times(2a-b)$
$(2a\times a)-(a\times b)+(2a\times b)-2(b\times b)$ and since a vector cross product itself zero.
=$a\times b$
but I am not sure if this be right.
I do not understand what you did. For 1. you just need to expand the product like it were an ordinary one: \begin{equation} (a+2b) \cdot(2a-b)=a\cdot 2a- a\cdot b+ 2b\cdot 2a -2 b\cdot b =2(||a||^2-||b||^2)+3a\cdot b \end{equation} For 2., again expanding the product and using the fact that for any two vectors $a$, $b$ one has $a\wedge a=0$, $a\wedge b=-b\wedge a$ \begin{equation} (a+2b)\wedge(2a−b)=0-a\wedge b+4b\wedge a+0 =-5a\wedge b \end{equation}