Question

Let \(S\)be the triangle with vertices \(\left( {4.2,1.2,4} \right)\), \(\left( {6,4,2} \right)\), \(\left( {2,2,6} \right)\). Find the image of Sunder the perspective projection with center of projection at \(\left( {0,0,10} \right)\).

Short answer

The coordinates of the vertices of the triangle are \(\left( {7,2,0} \right)\), \(\left( {7.5,5,0} \right)\), and \(\left( {5,5,0} \right)\).

Step-by-step solution

State the homogeneous coordinate of the vector

For the vector\(\left( {x,y,z} \right)\), thehomogeneous coordinates are\(\left( {x,y,z,1} \right)\).

Generally, for the vector\(\left( {x,y,z} \right)\), thehomogeneous coordinates are\(\left( {X,Y,Z,H} \right)\), where\(H \ne 0\).

The vector entries can be obtained as shown below:

\(x = \frac{X}{H}\), \(y = \frac{Y}{H}\), and \(z = \frac{Z}{H}\)

Obtain the homogeneous coordinates of the vertices of the triangle

Let Pbe the matrix that performs a projectionwith the center \(\left( {0,0,10} \right)\), as shown below:

\(P = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&{ - .1}&1\end{array}} \right]\)

The given vertices of the triangle \(S\) are \(\left( {4.2,1.2,4} \right)\),\(\left( {6,4,2} \right)\),\(\left( {2,2,6} \right)\).

Thehomogeneous coordinates of the vertices of the triangle are\(\left( {4.2,1.2,4,1} \right)\),\(\left( {6,4,2,1} \right)\),\(\left( {2,2,6,1} \right)\).

Thedata matrix can be represented as

\[\left[ {\begin{array}{*{20}{c}}{4.2}&6&2\\{1.2}&4&2\\4&2&6\\1&1&1\end{array}} \right]\].

Compute the data matrix

Obtain thedata matrix for the transformed triangle, as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&{ - .1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{4.2}&6&2\\{1.2}&4&2\\4&2&6\\1&1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4.2}&6&2\\{1.2}&4&2\\0&0&0\\{.6}&{.8}&{.4}\end{array}} \right]\)

Thus, the homogenous coordinates are \(\left( {4.2,1.2,0,.6} \right)\), \(\left( {6,4,0,.8} \right)\), and \(\left( {2,2,0,.4} \right)\).

Obtain the vector that has homogeneous coordinates

Compare thehomogeneous coordinates\(\left( {4.2,1.2,0,.6} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 4.2\),\(Y = 1.2\),\(Z = 0\), and\(H = .6\).

Now, obtain the vector entries as shown below:

\(\begin{array}{l}x = \frac{{4.2}}{{.6}}\\x = 7\end{array}\),

\(\begin{array}{l}y = \frac{{1.2}}{{.6}}\\y = 2\end{array}\),

And

\(\begin{array}{l}z = \frac{0}{{.6}}\\z = 0\end{array}\)

Thus, the coordinates are\(\left( {7,2,0} \right)\).

Compare thehomogeneous coordinates\(\left( {6,4,0,.8} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 6\),\(Y = 4\),\(Z = 0\), and\(H = .8\).

Now, obtain the vector entries as shown below:

\(\begin{array}{l}x = \frac{6}{{.8}}\\x = 7.5\end{array}\),

\(\begin{array}{l}y = \frac{4}{{.8}}\\y = 5\end{array}\),

and

\(\begin{array}{l}z = \frac{0}{{.8}}\\z = 0\end{array}\)

Thus, the coordinates are\(\left( {7.5,5,0} \right)\).

Compare thehomogeneous coordinates\(\left( {2,2,0,.4} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 2\),\(Y = 2\),\(Z = 0\), and\(H = .4\).

Now, obtain the vector entries as shown below:

\(\begin{array}{l}x = \frac{2}{{.4}}\\x = 5\end{array}\),

\(\begin{array}{l}y = \frac{2}{{.4}}\\y = 5\end{array}\),

And

\(\begin{array}{l}z = \frac{0}{{.4}}\\z = 0\end{array}\)

Thus, the coordinates are\(\left( {5,5,0} \right)\).

Therefore, the coordinates of the vertices of the triangle are \(\left( {7,2,0} \right)\), \(\left( {7.5,5,0} \right)\), and \(\left( {5,5,0} \right)\).