Solutions for Straight Line Representation

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Nonparametric equation 
for a line {x} through point with direction u:    
(− a 0. 

          This equation is saying, since a is on the line, x will be on the line if and only if vector xa is parallel to vector u.



Exercises:


          (1)  Sketch the line. 

                Here's an interactive diagram showing the line and bivectors mentioned in the other exercises below.
                Use the mouse to drag the point
P along the line. You can also drag the free copy of u around for comparison
                purposes. It doesn't change the picture at all whether we consider 
u to be a unit vector or not.




          (2)  Derive an equivalent parametric equation for the line with  u2 1.

                Hint:    Write  (xa)α  and solve for  x αa

                            Using the hint:

\[\Large{\begin{array}{l}0\,\, + \,\,\alpha \,\,\, = \,\,\,\left( {{\bf{x}}\,\, - \,\,{\bf{a}}} \right) \wedge {\bf{\hat u}}\,\, + \,\,\left( {{\bf{x}}\,\, - \,\,{\bf{a}}} \right) \cdot {\bf{\hat u}}\,\,\, = \,\,\,\left( {{\bf{x}}\,\, - \,\,{\bf{a}}} \right)\,\,{\bf{\hat u}}\\\,\,\,\,\,~~\,\alpha \,{\bf{\hat u}}\,\,\, = \,\,\,\left( {{\bf{x}}\,\, - \,\,{\bf{a}}} \right)\,\,{\bf{\hat u}}\,{\bf{\hat u}}\,\,\, = \,\,\,{\bf{x}}\,\, - \,\,{\bf{a}}\end{array}}\]


                
Or, more directly, since our line equation means vector xa is parallel to vector u, our last equation
                follows immediately.


          (3)  Find the directed distance from the origin to the line and sketch. 

                Hint:    xadu. Sketch the directed areas and solve for d.

                            See the interactive diagram above for the sketches. Our defining line equation shows that
                            
xau. That is, these bivectors are equivalent. Our interactive diagram above depicts them
                            as parallelograms; the blue one travels and changes parallelogram shape as the point moves
                            along the line, the green one is fixed. But, as we can also see from the diagram, we might as
                            well represent that bivector as a fixed (yellow) rectangle. Solving:

\[\Large{{\bf{d}}\,\,\, = \,\,\,{\bf{a}} \wedge {\bf{u}}\,\,{{\bf{u}}^{ - 1}}\,\,\, = \,\,\,{\bf{a}} \wedge {\bf{\hat u}}\,\,{\bf{\hat u}}}\]


          (4)  Find the directed distance from an arbitrary point to the line.

                Let's call this directed distance from an arbitrary point D. This amounts to calling the origin y instead
                of
O, so we can just slightly alter our last result:

\[\Large{{\bf{D}}\,\,\, = \,\,\,\left( {{\bf{a}}\,\, - {\bf{y}}\,} \right) \wedge {\bf{\hat u}}\,\,{\bf{\hat u}}}\]




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© David Hestenes 2005, 2014