Question

Let \(S\)be the triangle with vertices \(\left( {9,3, - 5} \right)\), \(\left( {12,8,2} \right)\), \(\left( {1.8,2.7,1} \right)\). 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 \(\left( {6,2,0} \right)\), \(\left( {15,10,0} \right)\), and \(\left( {2,3,0} \right)\).

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 projection with 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 triangle \(S\) are \(\left( {9,3, - 5} \right)\),\(\left( {12,8,2} \right)\), .

Thehomogeneous coordinates of the vertices of the triangle are\(\left( {9,3, - 5,1} \right)\),\(\left( {12,8,2,1} \right)\),\(\left( {1.8,2.7,1,1} \right)\).

Thedata matrix can be represented as

\[\left[ {\begin{array}{*{20}{c}}9&{12}&{1.8}\\3&8&{2.7}\\{ - 5}&2&1\\1&1&1\end{array}} \right]\].

Compute the data matrix

Obtain thedata matrix for the transformed triangle as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&{ - .1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}9&{12}&{1.8}\\3&8&{2.7}\\{ - 5}&2&1\\1&1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9&{12}&{1.8}\\3&8&{2.7}\\0&0&0\\{1.5}&{.8}&{.9}\end{array}} \right]\)

Thus, the homogenous coordinates are \(\left( {9,3,0,1.5} \right)\), \(\left( {12,8,0,.8} \right)\), and \(\left( {1.8,2.7,0,.9} \right)\).

Obtain the vector that has homogeneous coordinates

Compare thehomogeneous coordinates\(\left( {9,3,0,1.5} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 9\),\(Y = 3\),\(Z = 0\), and\(H = 1.5\).

Now, obtain the vector entries as shown below:

\(\begin{array}{l}x = \frac{9}{{1.5}}\\x = 6\end{array}\),

\(\begin{array}{l}y = \frac{3}{{1.5}}\\y = 2\end{array}\),

And

\(\begin{array}{l}z = \frac{0}{{1.5}}\\z = 0\end{array}\)

Thus, the coordinates are\(\left( {6,2,0} \right)\).

Compare thehomogeneous coordinates\(\left( {12,8,0,.8} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 12\),\(Y = 8\),\(Z = 0\), and\(H = .8\).

Now, obtain the vector entries as shown below:

\(\begin{array}{l}x = \frac{{12}}{{.8}}\\x = 15\end{array}\),

\(\begin{array}{l}y = \frac{8}{{.8}}\\y = 10\end{array}\),

And

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

Thus, the coordinates are\(\left( {15,10,0} \right)\).

Compare thehomogeneous coordinates\(\left( {1.8,2.7,0,.9} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 1.8\),\(Y = 2.7\),\(Z = 0\), and\(H = .9\).

Now, obtain the vector entries as shown below:

\(\begin{array}{l}x = \frac{{1.8}}{{.9}}\\x = 2\end{array}\),

\(\begin{array}{l}y = \frac{{2.7}}{{.9}}\\y = 3\end{array}\),

And

\(\begin{array}{l}z = \frac{0}{{.9}}\\z = 0\end{array}\)

Thus, the coordinates are\(\left( {2,3,0} \right)\).

Therefore, the coordinates of the vertices of the triangle are \(\left( {6,2,0} \right)\), \(\left( {15,10,0} \right)\), and \(\left( {2,3,0} \right)\).