# Orthogonal projection of a point on a line in 3D

Given a point A( 3,2,0) on a plane $$\alpha$$ = 2x+y-z-8=0 How to find the orthogonal projection of point A on a line r that has a direction vector ( 1,1,1) passes through the origin ? And What are the coordinates of A’ on r

• How is the plane relevant? – amd Jul 24 at 20:10

Hint: You need to find $$t$$ such that with $$A'=(t,t,t)$$ we have $$\vec{AA'} \perp \bf r$$, that is $$(3-t,\,2-t,\,-t)\cdot(1,1,1)=0$$.
• Do you see why $A'$ has to be in form $(t,t,t)$? – Berci Jul 24 at 11:36