The plane x/1+ y/2+ z/3= 1 intersect x–axis, y–axis, z–axis at A, B, C respectively. Find the distance between origin and orthocentre of triangle ABC.
My attempt: I found the triangle coordinates in 3D space but using direction cosine to find equation of line and then finding orthocentre seems pretty tough. Is there any other way to do this?
The solution states only one line ---orthocentre is the foot of perpendicular from origin on the plane. This seems completely weird to me and possible maybe but i have no idea how. Well I found this answer regarding a similar question but me being a high school student could anyone prove this without barycentric coordinates ? The link is as follows.
Coordinates of orthocentre,circumcentre and incentre of a triangle formed in 3d plane
That is a consequence of the Theorem of Three Perpendiculars.
Let $O$ be the origin: $AO$ is perpendicular to plane $OBC$ and, as a consequence of the above theorem, the projection $D$ of $O$ on line $BC$ is also the foot of the altitude $AD$ in triangle $ABC$.
On the other hand, if $H$ is the projection of $O$ on plane $ABC$, then by the same theorem $HD$ is perpendicular to line $BC$. It follows that $H$ lies on altitude $AD$.
But the argument can be repeated for the other vertices of $ABC$, hence $H$ also lies on the other altitudes of the triangle, and is therefore the orthocentre of $ABC$.