Q.The foot of the perpendicular drawn from the origin to a plane is M(1,0,0). Find the vector equation of the plane.

[!TLDR] When the foot of the perpendicular from the origin is $M$, the vector $\overrightarrow{OM}$ is itself the normal, and $M$ is also the point the plane passes through; apply $\vec r\cdot\vec n=\vec a …

The foot of the perpendicular from the origin is $M(1,0,0)$, so $\overrightarrow{OM} = \hat i$ is normal to the plane (the segment from the origin to the foot of the perpendicular is always perpendicular to the plane).

The plane passes through the point with position vector $\vec a = \hat i$, and is normal to $\vec n = \hat i$ (the same vector, since $M$ doubles as both). …

1. Recognise that when the foot of the perpendicular from the origin, $M$, is given, $\overrightarrow{OM}$ is a normal to the plane and $M$ is a point on the plane (this mirrors the worked Ex.(3) in this section). …

Treating $M(1,0,0)$ as an arbitrary point on the plane rather than realising $\overrightarrow{OM}$ is the normal too, and then being unable to proceed without another given vector; also mixing this up with the normal-form technique (Theorem 6.12) which needs $p$ and $\hat n$ rat …