# Computing dot products of linear combinations of unit vectors.

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 ?

• It is given that they are unit vectors, no specific values are mentioned though Commented Oct 9, 2014 at 18:38

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*}

• really nice, so b and c evaluate to $1-c^2$ thank you so much :) one stone two birds ! xD Commented Oct 9, 2014 at 18:41

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.

• could you please kindly elaborate a bit more Commented Oct 9, 2014 at 18:35
• thank you :) should it be okay to expand it using FOIL ? Commented Oct 9, 2014 at 18:39

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.