Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$

Question

Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$

The equation represents the line of intersection of two planes. Using augmented matrix $$ \begin{bmatrix} 1 & 1 & 1 & 3 \\ 2 & 3 & 4 & 6 \end{bmatrix}$$

$R_2\rightarrow R_2-2R_1$ $$ \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 0 \end{bmatrix}$$

$$y+2z=0$$ $$x+y+z=3$$ Equation of line is $$\frac{x-0}{1}=\frac{y-0}{-2}=\frac{z-0}{1}$$

Angle made by line and normal is $$\cos(90-\theta)=\frac{1}{\sqrt{6}}$$ where $\theta$ is the angle made by line and plane.

How should I proceed?

Answer 1

Note:

I get $$ y + 2z = 0 \Rightarrow z = - y/2 \\ 3 + z = (x + y + z) + z = x + y + 2z = x $$ So $$ \frac{x - 3}{1} = \frac{y-0}{-2} = \frac{z - 0}{1} $$ or $$ (x, y, z) = (3,0,0) + t (1, -2, 1) $$ Hint: The projection onto $z=0$ should affect what coordinate?

Spoiler:

The $z$-coordinate is forced to $0$. So we have $(x,y,z) = (3,0,0) + t (1,-2,0)$. We have $y = - 2 t$ and $x - 3 = t$ so $y = -2(x-3) = -2x + 6$ is the equation of the projected line within the $x$-$y$-plane.

Answer 2

A line passes through two points.

You can get the first point intersecting your line with the plane $z=0$. $p=(3,0,0)$.

Simply project another random point of your line, say $(0,6,-3)$, in the plane $z=0$, you get $q=(0,6,0)$.

The line you are looking for is thus $z=0$, $y=-2x+6$.