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Q. The projection of point $P (\vec{ p })$ on the plane $\vec{ r } \cdot \vec{ n }= q$ is $(\vec{ s })$, then

Three Dimensional Geometry

Solution:

We have $\vec{ s }-\vec{ p }=\lambda \vec{ n }$ and $\vec{ s } \cdot \vec{ n }= q$.
Thus, $(\lambda \vec{n}+\vec{p}) \cdot \vec{n}= q$
or $\lambda=\frac{ q -\vec{ p } \cdot \vec{ n }}{|\vec{ n }|^{2}}$
$\Rightarrow \vec{ s }=\vec{ p }+\frac{( q -\vec{ p } \cdot \vec{ n }) \vec{ n }}{|\vec{ n }|^{2}}$