Distance and locus

A line in space passes though the origin and forms equal angles with the axis. If the line intersects the plane $3x+5y+2z=30$ at the point ${\rm P}$ then:

1. Evaluate the distance between ${\rm P}$ and ${\rm O}$ , where ${\rm O}$ is the origin.
2. Find the equations of the locus of the points ${\rm M}$, for which the relation: $$\left | \overrightarrow{{\rm OM}} \right |= \left | \overrightarrow{\rm {PM}} \right |=4$$ holds. 

Solution

1. The line passes through the origin ${\rm O}$ and forms equal angles with the axis. The vector $\mathbf{a}=\left ( \cos \varphi, \cos \varphi, \cos \varphi \right )$ (where $\cos \varphi$ is the cosine of direction) , is a vector parallel to the line and for that it holds that:
   
   $$\cos^2 \varphi + \cos^2 \varphi + \cos^2 \varphi =1 \Leftrightarrow \cos \varphi = \frac{\sqrt{3}}{3}$$
   
   Hence $\mathbf{a}= \left ( \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3} \right )$ and thus the line has an equation $x=y=z$. Since the line intersects the plane at ${\rm P}$ a direct calculation reveals that ${\rm P}=(3, 3, 3)$. Thus the distance between ${\rm P}$ and the origin is:
   
   $${\rm d} \left ( {\rm P}, {\rm O} \right )= \sqrt{\left ( 3-0 \right )^2+ \left ( 3-0 \right )^2 + \left ( 3-0 \right )^2}=3\sqrt{3}$$
2. Let ${\rm M}$ be a point of the locus. Then $\left | \overrightarrow{{\rm OM}} \right |= \left | \overrightarrow{\rm {PM}} \right |=4$. This in return means:
   
   $$\begin{matrix}
   \sqrt{x^2+y^2+z^2}=4 & \\\\
    \sqrt{\left ( x-3 \right )^2+\left ( y-3 \right )^2+ \left ( z-3 \right )^2}=\sqrt{x^2+y^2+z^2}=4&
   \end{matrix}$$
   
   Simplifying we get that the requested locus is a circle, since we have an interesction of a sphere and of  a plane.