Find a orthogonal projection

Question

How to find a orthogonal projection of point $A(1,-2,3)$ in the plane $2x-3y+4z-6=0.$ Please help me. Thanky very much.

Answer 1

The direction vector of the line $AA′$ is $s = N = 2i -3 j + 4k,$ so the parametric equation of the line which is perpendicular to the plane and passes through the given point $A$

$$\frac{x-1}{2}=t\Rightarrow x=2t+1$$ $$\frac{y+2}{-3}=t\Rightarrow x=-3t-2$$ $$\frac{x-3}{4}=t\Rightarrow x=4t+3$$

these coordinates of the radius vector of the point $A′$ must satisfy the equation of the given plane that is

$$2(2t+1)-3(-3t-2)+4(4t+3)-6=0\Rightarrow t=-\frac{14}{29}$$

therefore, the coordinates of the point A′ are,

$x=2\cdot (-\frac{14}{29})+1=\frac{1}{29},y=\frac{-16}{29}, z=\frac{31}{29}$

thus the orthogonal projection of the point A onto the given plane is $A′(\frac{1}{29}, -\frac{16}{29}, \frac{31}{29})$

Answer 2

Say

$${A^*} = A + \lambda \cdot n$$

is the point your looking for.

$$n = \left( {2, - 3,4} \right)$$

is normal vector. Then:

$$\begin{gathered} \left\langle {{A^*},n} \right\rangle = 6 \hfill \\ \left\langle {A + \lambda \cdot n,n} \right\rangle = 6 \hfill \\ \left\langle {A,n} \right\rangle + \lambda \cdot \left\langle {n,n} \right\rangle = 6 \hfill \\ \lambda \cdot \left\langle {n,n} \right\rangle = 6 - \left\langle {A,n} \right\rangle \hfill \\ \lambda = \frac{{6 - \left\langle {A,n} \right\rangle }}{{\left\langle {n,n} \right\rangle }} \hfill \\ \end{gathered} $$

And

$${A^*} = A + \frac{{6 - \left\langle {A,n} \right\rangle }}{{\left\langle {n,n} \right\rangle }} \cdot n$$

For me $$\lambda = - \frac{{14}}{{29}}$$

$${A^*} = \frac{1}{{29}}\left( {1, - 16,31} \right)$$

Bracket-symbol means dot-product or scalar-product.