For any unit vectors $v$ and $w$, find the dot products (actual numbers) of:
a) $v$ and $-v$
b) $v+w$ and $v-w$
c) $v-2w$ and $v+2w$
I have worked part a :
a) $(v) \cdot (-v) = \cos (180) = -1$
Not getting any ideas on how to work part b and c... Any help ?
Dot product abide by distributive and commutative law. Feel free to distribute and commute.
b) $$ (v+w).(v-w) = v.v + v.(-w) + w.v + w.(-w) = 1 - v.w + v.w -1 = 0 $$ c) $$ (v-2w).(v+2w) = -3 \; \; \; (similarly) $$
Use the fact that dot products distribute over addition and dot products are commutative and scalars can be pulled out from either vector. For any scalar $c$, we have that: \begin{align*} (\vec v + c \vec w) \cdot (\vec v - c \vec w) &= \vec v \cdot \vec v - c(\vec v \cdot \vec w) + c(\vec v \cdot \vec w) - c^2(\vec w \cdot \vec w) \\ &= |\vec v|^2 - c^2|\vec w|^2 \\ \end{align*}
How about using the linearity of dot product? For the first component, we have $(a+b,c)=(a,c)+(b,c)$. Similarly for the second component.
Since the distributive and commutative properties apply to dot products (as said in other answers), the difference of two squares identity holds for dot product.
b)
$$(\mathit{v}+\mathit{w}) \cdot (\mathit{v}-\mathit{w}) = \mathit{v} \cdot \mathit{v} - \mathit{w} \cdot \mathit{w}=1-1=0$$
c)
$$(\mathit{v}-2\mathit{w}) \cdot (\mathit{v}+2\mathit{w}) = \mathit{v} \cdot \mathit{v} - 2\mathit{w} \cdot 2\mathit{w} = \mathit{v} \cdot \mathit{v} - 4(\mathit{w} \cdot \mathit{w})=1-4=-3 $$
For clarity, $\mathit{v} \cdot \mathit{v} = 1$ because $\mathit{v}$ is a unit vector (same for $\mathit{w}$) and $(c_1\mathit{v}) \cdot (c_2\mathit{w}) = c_1c_2(\mathit{v} \cdot \mathit{w})$ because scalar multiplication applies to dot product.
