A dot product question 
Let $a=4j+5k$ and $b=-5i+5j+2k$. Find $a\cdot(a\cdot b).$

And I tried to compute $(a.b).a$ as $$(0i+20j+10k)\cdot(4j+5k)=80j+50k$$  but this answer is wrong. So what wrong in my calculation?
 A: You compute the dot product which is a scalar
$$ a \cdot b = \pmatrix{a_x \\ a_y \\ a_z} \cdot \pmatrix{b_x \\ b_y \\ b_z} = a_x b_x + a_y b_y + a_z b_z $$
and multiply with the vector
$$ ( a\cdot b) a = (a_x b_x + a_y b_y + a_z b_z) \pmatrix{a_x \\ a_y \\ a_z} = \pmatrix{a_x (a_x b_x + a_y b_y + a_z b_z)\\ a_y (a_x b_x + a_y b_y + a_z b_z)\\ a_z (a_x b_x + a_y b_y + a_z b_z)} $$
A: $$a\cdot b=20+10=30\\a(a\cdot b)=30(4j+5k)=120j+150k$$
A: Assume that $i$, $j$, $k$ are standard orthonormal basis vectors for $\mathbb{R}^3$. Given $a=4j+5k$ and $b=-5i+5j+2k$, find $a\cdot (a\cdot b)$. 
If you meant $a\cdot (a\cdot b)$ where $\:\cdot\:$ is defined to be the dot product of two vectors, then the resulting expression (answer) should be $\textit{not possible}$ or $\textit{no solution}$ since you cannot take the dot product between a vector and a scalar (the expression $a\cdot b$ results in a scalar). 
However, if you meant $a(a\cdot b)=(a\cdot b)a$, then the resulting expression is 
$$
\begin{align*}
(a\cdot b)a &= ((4j+5k )\cdot (-5i+5j+2k))\langle 0,4,5\rangle \\ 
&= (\langle 0,4,5\rangle \cdot \langle -5,5,2 \rangle) \langle 0,4,5\rangle \\  
&= (20+10) \langle 0,4,5\rangle \\ 
&= 30 \langle 0,4,5\rangle \\  
&= \langle 0,120,150\rangle \\  
&= 120 j + 150 k. \\ 
\end{align*}
$$
A: You have to tell us what $i,j,k$ are. If they just make up the standard basis in $\mathbb{R^3}$ then they are orthonormal. So the dotproduct with themself is 1 and otherwise 0 (with the other unit vectors at least). 
Explicitly in your example $a\cdot b$ will be 30. Please review that the dotproduct takes two vectors and outputs a number.
