Problem based on area projection in 3-D geometry

Question

Considering a Quadrilateral $ABCD$ where $A(0,0,0), B(2,0,2), C(2,2\sqrt 2,2), D(0,2\sqrt2,0)$. Basically I have to find the Area of projection of quadrilateral $ABCD$ on the plane $x+y-z=3$.

I have tried to first find the projection of the points $A,B,C,D$ individually on the plane and then using the projected points find the vectors $\vec{AB}$ and $\vec{BC}$ and then using $|\vec{AB}\cdot \vec{BC}|$ , but I was unable to find the projected points.

Is it the correct approach? If it is not I would highly appreciate a correct approach for the problem.

Answer 1

As indicated by Intelligenti pauca in the above comments, the best way to go is to find the area of the quadrilateral, then multiply the area found by the cosine of the angle between the two planes which is the same angle between the normals to the planes (or its supplement).

$\begin{equation} \begin{split} \text{Area} &= \frac{1}{2} ( | AB \times AC | + | AC \times AD | ) \\ &= \frac{1}{2} ( | (2,0,2) \times (2, 2\sqrt{2}, 2) | + | (2, 2 \sqrt{2}, 2) \times (0, 2 \sqrt{2}, 0) | ) \\ &= 2 ( | (- \sqrt{2}, 0, \sqrt{2} ) | + | (-\sqrt{2}, 0, \sqrt{2} ) | )\\ & = 2 ( 2 + 2 ) = 8 \end{split} \end{equation}$

The normal to the plane of the quadrilateral is along $(-1, 0, 1)$ and the normal to the projection plane is along $(1, 1, -1)$, therefore, if $\theta$ is the angle between these two normals, then,

$\cos \theta = \dfrac{ (-1, 0, 1) \cdot (1, 1, -1) }{\sqrt{2}\sqrt{3} } = -\sqrt{\dfrac{2}{3}} $

Hence, the projected area is equal to $8\sqrt{\dfrac{2}{3}}$

Answer 2

You can also first find the projection of points and then calculate the area.

Hints for finding the projection of a point:

$x+y-z=3$

So the normal vector is $N(1, 1, -1)$

the projection of any point on this plane is the intersection of plane and a line passing that point and perpendicular on plane, that is the gradients of line is:

$(m, n, l)=(1, 1,-1)$

For example we find projection of point A:

equation of line passing A and perpendicular on plane is:

$\frac{x-0}m=\frac{y-0}n=\frac{z-0}l$

$\frac x1=\frac y1=\frac z{-1}=t$

$\Rightarrow x=t, y=t, z=-t$

Now plug in these in equation of plane:

$t+t-(-t)=3\Rightarrow t=1$

So the coordinates of A' (projection of A) is:

$x=t=1$, $y=t=1$ and $z=-t=-1$ or $A'(1, 1, -1)$

Now find B', C' and D' and use similar method given in other answer to find the area of A'B'C'D'.