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

  • $\begingroup$ How is the plane relevant? $\endgroup$ – 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 $(3-t,\,2-t,\,-t)\cdot(1,1,1)=0$.

  • $\begingroup$ -3t+5=0 —> t=5/3 thank you so much $\endgroup$ – Tarek Chafei Jul 24 at 11:34
  • $\begingroup$ Do you see why $A'$ has to be in form $(t,t,t)$? $\endgroup$ – Berci Jul 24 at 11:36
  • $\begingroup$ Because A’ is a point on a line r, then it has the components of a generic point ( x,y,z ) of r which is ( x=t,y=t,z=t ) ? $\endgroup$ – Tarek Chafei Jul 24 at 11:41

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