Question

What vector in \({\mathbb{R}^{\bf{3}}}\) has homogeneous coordinates \(\left( {\frac{1}{2}, - \frac{1}{4},\frac{1}{8},\frac{1}{{24}}} \right)\)?

Short answer

The vector is \(\left( {12, - 6,3} \right)\).

Step-by-step solution

State the homogeneous coordinate of the vector

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 vector that has homogeneous coordinates

Compare thehomogeneous coordinates\(\left( {\frac{1}{2}, - \frac{1}{4},\frac{1}{8},\frac{1}{{24}}} \right)\)with the generalhomogeneous coordinates \(\left( {X,Y,Z,H} \right)\)to get\(X = \frac{1}{2}\),\(Y = - \frac{1}{4}\),\(Z = \frac{1}{8}\), and\(H = \frac{1}{{24}}\).

Now, obtain the vector entries, as shown below:

\(\begin{array}{l}x = \frac{{1/2}}{{1/24}}\\x = 12\end{array}\),

\(\begin{array}{l}y = \frac{{ - 1/4}}{{1/24}}\\y = - 6\end{array}\),

And

\(\begin{array}{l}z = \frac{{1/8}}{{1/24}}\\z = 3\end{array}\)

Thus, the vector is \(\left( {12, - 6,3} \right)\).