91Ó°ÊÓ

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 \(z = \frac{Z}{H}\)

Compare the givenhomogeneous coordinates\(\left( {1, - 2,3,4} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 1\),\(Y = - 2\),\(Z = 3\), and\(H = 4\).

Now, obtain vector entries, as shown below:

\(x = \frac{1}{4}\),

\(\begin{array}{l}y = \frac{{ - 2}}{4}\\y = - \frac{1}{2}\end{array}\)

And

\(z = \frac{3}{4}\)

Thus, the point is\(\left( {\frac{1}{4}, - \frac{1}{2},\frac{3}{4}} \right)\).

Compare thehomogeneous coordinates\(\left( {10, - 20,30,40} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = 10\),\(Y = - 20\),\(Z = 30\), and\(H = 40\).

Now, obtain the vector entries, as shown below:

\(\begin{array}{l}x = \frac{{10}}{{40}}\\x = \frac{1}{4}\end{array}\)

\(\begin{array}{l}y = \frac{{ - 20}}{{40}}\\y = - \frac{1}{2}\end{array}\)

And

\(\begin{array}{c}z = \frac{{30}}{{40}}\\ = \frac{3}{4}\end{array}\)

Thus, the point is \(\left( {\frac{1}{4}, - \frac{1}{2},\frac{3}{4}} \right)\).

From the above result, thehomogenous coordinates \(\left( {1, - 2,3,4} \right)\)and\(\left( {10, - 20,30,40} \right)\)represent the same point\(\left( {\frac{1}{4}, - \frac{1}{2},\frac{3}{4}} \right)\).

Thus, the sets of homogeneous coordinates represent the same point in \({\mathbb{R}^3}\).