# Deriving Parallel and perpendicular vectors from triple vector product

How would one go about resolving the vector $$\vec{p}$$ into parallel and perpendicular vectors to the given vector $$\vec{w}$$

By considering - $$\vec{w}\times(\vec{p}\times\vec{w})$$

So far I have used the triple vector product however I seem to just get zero when I do this so I feel like i'm making a mistake somewhere.

Inasmuch as $$\vec p\times \vec w$$ is perpendicular to both $$\vec p$$ and $$\vec w$$, we can decompose $$\vec p$$ as

\begin{align} \vec p&=A\vec w+B[\vec w\times(\vec p\times \vec w)]\tag1 \end{align}

Note that $$\vec w\times(\vec p\times \vec w)$$ is perpendicular to $$\vec w$$.

Taking the inner product of $$\vec p$$ with $$\vec w$$, we find from $$(1)$$ that

$$A=\frac{\vec p\cdot \vec w}{|\vec w|^2}$$

Taking the vector product of $$\vec p$$ with $$\vec w$$, we find from $$(1)$$ that

$$B=\frac{1}{|\vec w|^2}$$

Hence, denoting the unit vector along $$\vec w$$ as $$\hat w=\frac{\vec w}{|\vec w|}$$

$$\vec p=(\vec p\cdot \hat w)\hat w+ ( \hat w \times\vec p)\times \hat w$$

• Ahhh I see, thank you – AH_01 Aug 3 '20 at 20:09