Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0
Answer:
(3,-4,2)
Hello
Let $\displaystyle H$ be the orthogonal projection of $\displaystyle A$ on the plane which equation is $\displaystyle x-3y-z=13$
We know two things about $\displaystyle H$ :
- As the projection is orthogonal, $\displaystyle \vec{AH}$ is orthogonal to the plane.
- $\displaystyle H(x_h,y_h,z_h)$ lies in the plane : $\displaystyle x_h-3y_h-z_h=13$
Luckily, the equation of the plane $\displaystyle {\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 : $\displaystyle \vec{n}=\begin{pmatrix}1\\-3\\-1\end{pmatrix}$. As $\displaystyle \vec{n}$ and $\displaystyle \vec{AH}$ are both orthogonal to the plane, they share the same direction that is to say that there exists a real number $\displaystyle t$ such that $\displaystyle \vec{AH}=t\vec{n}$.
As $\displaystyle \vec{AH}=\begin{pmatrix}x_h-2\\y_h-(-1)\\z_h-3 \end{pmatrix}$ and $\displaystyle t\vec{n}=\begin{pmatrix}t\\-3t\\-t \end{pmatrix}$ we have four equations :
$\displaystyle
\begin{cases}
t=x_h-2 \\
-3t=y_h+1\\
-t=z_h-3\\
13=x_h-3y_H-z_h
\end{cases}\implies \begin{cases}
x_h=t+2 & (1)\\
y_h=-3t-1& (2)\\
z_h=-t+3&(3)\\
13=x_h-3y_H-z_h&(4)
\end{cases}$
Substituting $\displaystyle (1),\,(2)$ and $\displaystyle (3)$ in $\displaystyle (4) $ will give you the value of $\displaystyle t$ from which you'll get $\displaystyle H(x_h,y_h,z_h)$. Good luck