A plane $\Pi$ has an equation of $\mathbf{r}\cdot\mathbf{n}=k$. If the origin is on the plane explai

Question

A plane $\Pi$ has an equation of $\mathbf{r}\cdot\mathbf{n}=k$. If the origin is on the plane explain how the value of $k$ could be determined and state its value. State the unit vectors of the normal of $x-y$, $y-z$ and $x-z$ planes. Thus, find the equations of each plane. Show that the shortest distance of plane $\Pi$ to the origin is $\frac{|k|}{|\mathbf{n}|}$.

Accepted Answer

(i) $k = 0$ (ii) $\Pi_{xz}: \mathbf{r} \cdot (0,1,0) = 0$; $\Pi_{yz}: \mathbf{r} \cdot (1,0,0) = 0$; $\Pi_{xy}: \mathbf{r} \cdot (0,0,1) = 0$ (iii) Shortest distance $= \frac{|k|}{|\mathbf{n}|}$

The x-y Plane, y-z Plane and x-z Plane Equation

The x-y Plane, y-z Plane and x-z Plane Equation

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