A light and a shadow

The shadow area $ABC'D '$ is created by projecting the rectangle $ABCD$ into the $xy$ -plane. The segements $\overline {CE}$ and $\overline {DE}$ between the light source and the corners of the billboard can be extended to the lines $g$ and $h$ : $\begin{aligned} g: \vec E + \lambda \cdot (\vec C - \vec E) &= \left(\!\begin{array}{c} 3 \\ -1 \\ 4 \end{array} \!\right) + \lambda \cdot \left(\!\begin{array}{c} -3 \\ 4 \\ -2 \end{array} \!\right) \\ h: \vec E + \mu \cdot (\vec D - \vec E) &= \left(\!\begin{array}{c} 3 \\ -1 \\ 4 \end{array} \!\right) + \mu \cdot \left(\!\begin{array}{c} -3 \\ 1 \\ -2 \end{array} \!\right) \end{aligned}$ where $\lambda,\mu \in \mathbb{R}$ are scalars. For the parameters $\lambda = \mu = 2$ the straight lines pierce the $xy$ -plane. The points of the shadow area then result $\begin{aligned} \vec C' &= \left(\!\begin{array}{c} 3 \\ -1 \\ 4 \end{array} \!\right) + 2 \cdot \left(\!\begin{array}{c} -3 \\ 4 \\ -2 \end{array} \!\right) = \left(\!\begin{array}{c} -3 \\ 7 \\ 0 \end{array} \!\right) \\ \vec D' &= \left(\!\begin{array}{c} 3 \\ -1 \\ 4 \end{array} \!\right) + 2 \cdot \left(\!\begin{array}{c} -3 \\ 1 \\ -2 \end{array} \!\right) = \left(\!\begin{array}{c} -3 \\ 1 \\ 0 \end{array} \!\right) \end{aligned}$ The shadow has the shape of a trapezium with the two side lengths $a = 3$ and $c = 6$ and the height $h = 3$ . The area then gives $A = \frac{1}{2} (a + c) h = 13.5$

The z-component of any point on the shadow will be 0. Let the point of the shadow of points $A, B, C, D$ be represented by $A', B', C', D'$

$A'$ and $B'$ are already on the xy-plane, so therefore their coordinates are same as $A$ and $B$ respectively -- ( $0, 0, 0)$ and ( $0, 3, 0)$

$C'$ and $D'$ can be found by calculating when line $EC = \begin{pmatrix} 0 \\ 3 \\ 2 \end{pmatrix} +\begin{pmatrix} 3 \\ -4 \\ 2 \end{pmatrix}t$ and line $ED = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} +\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}t$ intersect the xy-plane (i.e when z-component equals zero) In both cases, z = 0 when t = -1, and therefore $C' = (-3, 7, 0)$ and $D' = (-3, 1, 0)$

Ignoring the z component, I used the shoelace formula to calculate the area: $A = \frac{1}{2}|[(0)(3)+(0)(7)+(-3)(1)+(-3)(0)]-[(0)(0)+(3)(-3)+(7)(-3)+(1)(0)]| = \frac{1}{2}|-3 + 30| = \boxed{13.5}$

General solution, averages

Let $E\ (-x,y,z)$ , $C\ (0,w,h)$ , $D\ (0,0,h)$ , and the vertices of the shadow $C'\ (-s,q,0)$ , $D'\ (-s,p,0)$ .

Point $C$ lies on the line segments $EC'$ . This means that its coordinates are a weighted average of the coordinates of the endpoints, $C = \frac{\lambda E + \mu C'}{\lambda + \mu},\ \ \ \text{or}\ \ \ \ (\lambda + \mu)C = \lambda E + \mu C'.$ This leads to the system of equations $\begin{aligned} 0 & = \lambda x - \mu s \\ (\lambda + \mu) w & = -\lambda y + \mu q \\ (\lambda + \mu) h & = \lambda z. \end{aligned}$ The bottom equation is satisfied if $\lambda = h$ and $\lambda + \mu = z$ ; the top two equations become $\begin{aligned} 0 & = hx - (z-h) s \\ zw & = -hy + (z-h) q \end{aligned}$ showing that $s = \frac{h}{z-h} x,\ \ \ \ q = \frac{zw + hy}{z-h}.$ Likewise, point $D$ lies on the line segment $ED'$ . The analysis is the same, with $w$ replaced by $0$ : $p = \frac{h}{z-h} y.$

Thus the shadow is a trapezoid in the plane $z = 0$ with vertices $A\ (0,0)$ , $B\ (0,w)$ , $C\ (-s,q)$ , $D\ (-s,p)$ ; its bases are $b_1 = w$ , $b_2 = q - p$ , and its height $c = s$ . The area is $A = \tfrac12c(b_1 + b_2) = \tfrac12s(w + q - p) = \frac12\frac{hx}{z-h}\left(w + \frac{zw}{z-h}\right);$ $A = \frac{hxw(2z - h)}{2(z-h)^2}.$ (Note that $y$ dropped out when we subtracted $q - p$ .) Substituting $h = 2, x = 3, z = 4, w = 3$ we obtain $A = \frac{2\cdot 3\cdot 3\cdot 6}{2\cdot 2^2} = \frac{108}{8} = \boxed{13\tfrac12}.$

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General solution, similar triangles

Define $d = FG$ the distance of the light source to the plane of the rectangle; $h = HG = AD$ the height of the rectangle; $z = EF$ the height of the light source above the ground; $w = AB$ the width of the rectangle.

Use similar triangles to see that $\frac{FH'}{FG} = \frac{EF}{EF - HG} = \frac{z}{z - h} =: k,$ and that triangles $FD'C'$ is similar to $FAB$ with factor $k$ . Thus $\text{area}\ \triangle FD'C' = k^2\cdot \text{area}\ \triangle FAB,$ and $\begin{aligned} \text{shadow area} & = \text{area}\ \triangle FD'C' - \text{area}\ \triangle FAB \\ & = (k^2 - 1)\cdot \text{area}\ \triangle FAB \\ & = \frac{k^2-1}2 wd \end{aligned}.$ In this case, $d = 3$ , $h = 2$ , $z = 4$ , $w = 3$ , so we find $k = \frac{4}{4 - 2} = 2;$ $\text{shadow area} = \frac{2^2 - 1}2\cdot 3 \cdot 3 = \boxed{13\tfrac12}.$

Cleary A and B will stay where they are. We can find the images of C and D using vectors to project them to where Z=0. From E to C is <-3,4,-2> how convenient that this is the exact vector that brings C to C' at (-3,7,0). From E to D the vector is <-3,1,-2> and again this brings D to D' at (-3,1,0).

The quadrilateral ABC'D' is a trapezoid with height 3 (The distance in the x-direction from 0 to -3) and bases AB=3 and C'D'=6 (the difference in the y-direction from 1 to 7.)

$A=\frac{1}{2}h(b1+b2)=\frac{1}{2}*3*(3+6)=\boxed{13.5}$

Incidentally, the y-coordinate of the light is not important. The fact that the z-coordinate of E is twice that of C and D made the height of the trapezoid become the x-coordinate of E. The base of 6 comes from the x-coordinate of E being the opposite of those of C' and D'. Form triangle EC'D' and you can see CD is a midpoint connector.