There are various forms of the Cartesian equations of lines in 2D:
y = mx + c
ax + by + c = 0
y - y 1 = m(x – x1)
3D Cartesian equations of lines are much more complicated– 3D lines are easier to do in vectors, let's learn that in A-Level Maths.
Suppose you have two vectors in a plane:
We shall imagine vector a starting at the origin. That makes it into the position vector of the point at its other end.
Now look at a + b:
This gives the position vector of the point at the end of b.
Now for a + 2b:
Now a – b:
Notice that we’ve let the first two examples leave a trace.
Lastly, a –2b:
The aim of this demonstration is to show that, whatever multiple of b you add on to a, you will land up somewhere on the line through the end of a, whose direction is determined by b.
We therefore write the equation of this line as:
r = a + lamda x b
where lamda is a parameter which varies from - ♾ to + ♾
NB1
All the directional information is in b.
NB2
You could replace a by the position vector of any other point on the line, so this process does not produce a unique equation for the line.
NB3
Although we’ve looked at this as a 2D issue, there’s no difference at all in the equation if the vectors are 3D.
Example
Convert this to a Cartesian equation. This is done by eliminating lamda.
First, we need to spell out the three separate equations which are hidden within the vector equation. Understanding r to mean
we have three separate equations for the x, y and z components of the vectors:
x = 3 + lamda
y = 2 + 2lamda
z = 4 + 4lamda
Now we changethe subject of each of these to lamda:
These three are all for the same lamda, so we can put the three equal to each other. Without changing its meaning, we can reformat the x entry to be the same layout as the others:
This is the Cartesian equation of the line.
Notice that the denominators of the three fractions are the components of the direction vector– that is, they contain all the directional information for the line.
Also, the numbers subtracted from x, y and z on the numerators are the coordinates of a point on the line.
You might also notices that this equation is a lot more awkward to handle than y = mx+ c, its two-dimensional equivalent.
For myself, I only deal with 3D Cartesian equations by finding a vector equation and translating it into Cartesian.
N.B.
Any one line has infinitely many vector equations.
So any 3D line has infinitely many Cartesian equations.
Drafted by Eunice (Maths)