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.


1 Answer 1


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 $\hat w=\frac{\vec w}{|\vec w|}$

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

  • $\begingroup$ Ahhh I see, thank you $\endgroup$
    – AH_01
    Aug 3, 2020 at 20:09

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