Computing dot products of linear combinations of unit vectors.
Andrew Henderson 
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 ?
$\endgroup$ 14 Answers
$\begingroup$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) $$ (v-2w).(v+2w) = -3 \; \; \; (similarly) $$
$\endgroup$ $\begingroup$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*}
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
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