Projection of vector onto the plane. I have vector $b = (1, 1, 1)$ and the plane $2x - y + z = 1$.
I know how to find the projection of the vector on the vector, but how to do this with the plane?
 A: A plane is uniquely defined by a point and a vector normal to the plane. The equation of the plane $2x-y+z=1$ implies that $(2,-1,1)$ is a normal vector to the plane. If you project the vector $(1,1,1)$ onto $(2,-1,1)$, the component of $(1,1,1)$ that was "erased" by this projection is precisely the component lying in the plane. So, 
$$b- \text{proj}_{(2,-1,1)}(b) = \text{proj}_{2x-y+z=1}(b).$$
A: Guide for this particular question:


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*If a point lies on the plane, then its projection on the plane is the point itself.

A: $2x-y+z=1$ implies a normal to the plane of $<2,-1,1>$. So the unit normal is: $\hat{n}=\frac{1}{\sqrt{6}}<2,-1,1>$
You can subtract the projection of vector <1,1,1> onto the normal, alternatively, but equivalently, you can express this as a cross product: $\hat{n}\times(\vec{b}\times\hat{n})=\vec{b}-\hat{n}(\hat{n}\cdot \vec{b})$
Generally $\vec{b}=\hat{n}\times(\vec{b} \times\hat{n}) + \hat{n}(\hat{n}\cdot \vec{b}) $ for any unit vector $\hat{n}$.
This decomposition is often very useful in proving identities. 
