Difficulty solving problem regarding 2 parallel vectors I have been asked to solve the following question, and I am not sure where to begin with it : 
Given two vectors $a$ and $b$ , where the magnitude of $b$ is double the
magnitude of $a$ and the angle between the two vectors is $60°$, we may
define two further vectors: $c = a − 2b$ and $d =14a − 2b$ .
Show that the vectors c and d are perpendicular to each other.
I am aware that $$a\cdot b = \vert a\vert\cdot \vert b\vert\cdot cos $$
However im not sure how to apply it in this question.
Any help in solving this question and the methodology to solve it would be greatly appreciated

$$c\cdot d=(a-2b)\cdot (14a-2b)=14|a|^2-30a\cdot b+4|b|^2$$
Use that 
$$|b|=2|a| \quad (1)$$
and then
$$a\cdot b=|a||b|\cos 60º \rightarrow a\cdot b=2|a|^2\cdot1/2=|a|^2 \quad (2)$$
Remember that $c \perp d \Leftrightarrow c \cdot d=0$.
Using $(1)$ and $(2)$ we get:
$$c\cdot d=14|a|^2-30a\cdot b+4|b|^2=14|a|^2-30|a|^2+4\cdot 4|a|^2=0$$
