Thread: projection of point and plan

1. projection of point and plan

Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0

(3,-4,2)

2. Hello
Originally Posted by Apprentice123
Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0

(3,-4,2)
Let $H$ be the orthogonal projection of $A$ on the plane which equation is $x-3y-z=13$

We know two things about $H$ :
• As the projection is orthogonal, $\vec{AH}$ is orthogonal to the plane.
• $H(x_h,y_h,z_h)$ lies in the plane : $x_h-3y_h-z_h=13$

Luckily, the equation of the plane ${\color{red}1}\cdot x{\color{red}-3}\cdot y{\color{red}-1}\cdot z=13$ gives us a vector which is orthogonal to the plane : $\vec{n}=\begin{pmatrix}1\\-3\\-1\end{pmatrix}$. As $\vec{n}$ and $\vec{AH}$ are both orthogonal to the plane, they share the same direction that is to say that there exists a real number $t$ such that $\vec{AH}=t\vec{n}$.

As $\vec{AH}=\begin{pmatrix}x_h-2\\y_h-(-1)\\z_h-3 \end{pmatrix}$ and $t\vec{n}=\begin{pmatrix}t\\-3t\\-t \end{pmatrix}$ we have four equations :

$
\begin{cases}
t=x_h-2 \\
-3t=y_h+1\\
-t=z_h-3\\
13=x_h-3y_H-z_h
\end{cases}\implies \begin{cases}
x_h=t+2 & (1)\\
y_h=-3t-1& (2)\\
z_h=-t+3&(3)\\
13=x_h-3y_H-z_h&(4)
\end{cases}$

Substituting $(1),\,(2)$ and $(3)$ in $(4)$ will give you the value of $t$ from which you'll get $H(x_h,y_h,z_h)$. Good luck

3. Thank you . I do AH (x-2,y+1,z-3) = (1,-3,-1). And I find x, y and z.

Thank you very much