# Prove that $perp_\vec{n}(\vec{x})=\frac{\vec{x}\cdot\vec{a}}{||\vec{a}||^2}\vec{a}+\frac{\vec{x}\cdot\vec{b}}{||\vec{b}||^2}\vec{b}$

suppose that {$\vec{a},\vec{b}$} is a basis for a plane in $\Bbb{R^3}$ such that $\vec{a},\vec{b}$ are orthogonal and $\vec{n}=\vec{a}\times\vec{b}$ how would you show that for all $\vec{x}$, $\text{perp}_\vec{n}(\vec{x})=\frac{\vec{x}\cdot\vec{a}}{||\vec{a}||^2}\vec{a}+\frac{\vec{x}\cdot\vec{b}}{||\vec{b}||^2}\vec{b}$

I have been able to show that $\text{perp}_\vec{n}(\vec{x})$ does lie on the plane since it is orthogonal to the normal vector, but i have't been able to show $\text{perp}_\vec{n}(\vec{x})=\frac{\vec{x}\cdot\vec{a}}{||\vec{a}||^2}\vec{a}+\frac{\vec{x}\cdot\vec{b}}{||\vec{b}||^2}\vec{b}$

• Hint: show that $\;\vec{n} \times \left(\vec{x} - \text{perp}_\vec{n}(\vec{x})\right) = 0\,$.
– dxiv
Feb 3, 2018 at 18:52
• How would i do that, it makes sense that it would 0 Feb 3, 2018 at 18:54

Hint: what needs to be shown is that $\,\vec{x} - \text{perp}_\vec{n}(\vec{x})\,$ is collinear with $\,\vec{n}\,$ i.e. the cross product is $\,0\,$:

\begin{align} \left(\vec{x} - \text{perp}_\vec{n}(\vec{x})\right) \times \vec{n} &= \vec{x} \times\vec{n} - \frac{\vec{x}\cdot\vec{a}}{||\vec{a}||^2}\vec{a}\times\vec{n}-\frac{\vec{x}\cdot\vec{b}}{||\vec{b}||^2}\vec{b}\times\vec{n} \tag{1} \end{align}

Using that $\,\vec{n}=\vec{a}\times\vec{b}\,$ and the triple product identity $\,\vec{u}\times(\vec{v}\times\vec{w})=(\vec{u}\cdot\vec{w})v-(\vec{u}\cdot\vec{v})w\,$ identity:

$$\vec{x}\times\vec{n} = \vec{x} \times (\vec{a}\times \vec{b}) = (\vec{x}\cdot\vec{b})\vec{a}-(\vec{x}\cdot\vec{a})\vec{b}$$

Then the RHS of $(1)$ becomes:

$$\require{cancel} (\bcancel{\vec{x}\cdot\vec{b})\vec{a}}-\cancel{(\vec{x}\cdot\vec{a})\vec{b}} - \frac{\vec{x}\cdot\vec{a}}{||a||^2}\left((\vec{a}\cdot\vec{b})\vec{a}-\cancel{||a||^2\vec{b}}\right) - \frac{\vec{x}\cdot\vec{b}}{||b||^2}\left(\bcancel{||b||^2 \vec{a}} - (\vec{a}\cdot\vec{b})\vec{b}\right) = \ldots$$

[ EDIT ]   With $\,\vec{u}=\frac{\vec{a}}{||a||}\,$, $\,\vec{v}=\frac{\vec{b}}{||b||}\,$, $\,\vec{w}=\frac{\vec{n}}{||n||}\,$, the vectors $\vec{u}, \vec{v}, \vec{w}$ form an orthonormal basis of $\,\mathbb{R}^3\,$ so it is "obvious" that any vector $\,\vec{x}\,$ decomposes as $\,\vec{x}=(\vec{x}\cdot\vec{u})\vec{u}+(\vec{x}\cdot\vec{v})\vec{v}+(\vec{x}\cdot\vec{w})\vec{w}\,$, from which the original proposition follows. The above is just a way to prove this intuition from first principles.