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Identifying Lines in Space

Instructor: Nadia Beherens

I have worked at several schools and universities as a Math teacher. I have a degree in Math teaching, and I am a master student.

In this lesson we will define the vector, parametric, continuous (or symmetric) and implicit equations of the line in space, and we will solve a problem.

There are lines everywhere around us; on the edge of a sheet of paper, for example, or on the outline of a table. Mathematically, a line is defined as a line of points that extends infinitely in two directions. The first examples, formally, are line segments -- that is, a part of a line that has defined endpoints -- because true lines are very hard to find in real life. They are, however, highly useful in mathematics. There are several ways to represent the location of a line in space.

1.) Vector equation: We need a point of the line and a direction vector, which gives the direction of the line. In the image, P( x0 , y0 , z0 ) can be the point and d = ( d1 , d2 , d3 ) the director vector.

A line portrayed on 3D graph, passing through a designated point and parallel to an arrow indicating a direction vector

So, the line r is:

r : (x , y , z) = ( x0 , y0 , z0 ) + t ( d1 , d2 , d3 )

How to calculate a direction vector?

If we have 2 points of the line, P( x0 , y0 , z0 ) and Q( x1 , y1 , z1 ), we can calculate the direction vector by subtracting the coordinates of either point from those of the other, using the general equation:

d = PQ = (x0-x1,y0-y1,z0-z1

For example:

If we have two points of the line, P=(4,7,2) and Q=(3,-1,4):

d = PQ = (3-4,-1-7,4-2)

d=(-1,-8,2)

Then, the equation of the line will be:

(x, y, z) = (4,7,2) + t(-1,-8,2)

2.) Parametric equations, which can be derived from the vector equation. Operating on the vector line, we obtain the following equality:

r : (x , y , z) = ( x0 + t d1 , y0 + t d2, z0 + t d3)

Then,

x = x0 + t d1

y = y0 + t d2 This set of equations is the parametric form of the equation.

z = z0 + t d3

3.) The continuous or symmetric equation, which can be derived from the parametric equation. Isolating t in each equation, we get:

t=x-x0/d1

t=y-y0/d2

t=z-z0/d3

Then, if we equate t, we obtain:

x-x0 over d1=y-y0 over d2=z-z0 over d3

Which is the continuous or symmetric equation.

We can find a point on a line defined by any of these equations by replacing t with any value.

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