# Vector magnitude and dot products

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?

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