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Homework Statement
Find vector \overrightarrow{c} collinear with the vector \overrightarrow{a}+\overrightarrow{b}, if \overrightarrow{a} \cdot \overrightarrow{b}=5 and \overrightarrow{c} \cdot \overrightarrow{b} = 18, |\overrightarrow{b}|=2
Homework Equations
|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}|sin(\overrightarrow{a},\overrightarrow{b})
\mathbf{a}\times\mathbf{b}=\det \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{bmatrix}.
\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}|cos(\overrightarrow{a},\overrightarrow{b})
The Attempt at a Solution
Before I started I found some (I believe) error in the task.
If I subtract the given equations like a.b-c.b=-13 and b(a-c)=-13
|b||a-c|cos(b, a-c) = -13
As we can see on the picture a-c0=-b
|b||a-pc0|cos(b,a-pc0)=-13
Edit: Ok, there isn't error I mixed c0 and c.
Now, let me start.
I approach
c = p(a+b), because c and a+b are collinear, where p is real number.
From the given conditions.
|a||b|cos(a,b)=5
|c||b|cos(c,b)=18
|a|cos(a,b)=5/2
|c|cos(c,b)=18/2
c.(a+b)=|c||a+b|cos(c,a+b)=|c||c0|cos(c,c0)=|c|
Here is where I am stuck...

II approach
c x (a+b) = 0
c x a + c x b =0
But as you can see I got no coordinates for the vectors, so this is 2nd fail...

I am dealing with this task for 1 hour and seems like I can't find any way to solve it.
The result in the textbook is c = 2(a+b)
Thanks in advance.
Regards.
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