Vector magnitude and dot products

Question

Assume that $v, w \in R$ and $∥v∥ = 2$, $∥w∥ = 3$ and $v \cdot w = -1$

Let $a = 3v−w$ and $b = v+w$. Determine $a \cdot b$

How should one approach this? I could find the vectors $v$ and $w$ if i knew the angle between them since i have their magnitudes and dot product right? The dot product is -1, isn't the angle 180 degrees then?

Answer 1

$$ a\cdot b = (3v-w)\cdot(v+w) = 3 v\cdot v + 3 v\cdot w - w\cdot v - w\cdot w = 3\|v\|^2 + 2 v\cdot w - \|w\|^2 = 12 - 2 - 9 = 1 $$

Answer 2

$(3v-w)\cdot (v+w)=3v\cdot v+3 v\cdot w-w\cdot v -w\cdot w= 3\|v\|^{2} +2 v\cdot w -\|w\|^{2}=12-2-9=1$.

Answer 3

Other answers already tell you to just write down the dot product $a\cdot b$, substitute to get an expression in $v$ and $w$, and use bilinearity.

Your approach to first determine $v$ and $w$ doesn't work, because the information given doesn't uniquely determine $v$ and $w$. For example, in the real plane, $(3,0)$ and $(-1/3,\sqrt{35}/9)$ are possible solutions for $v$ and $w$ (I found this by writing $v=(3,0)$ and solving for $w=(x,y)$), but you can rotate these solutions over the origin to get other pairs of solutions. Also note that while $v\cdot w =-1 $, the vectors aren't linearly dependent. That only follows when their norms are both equal to $1$.