Scalar Equations!

[Saad]

Scalar equations such as y=2x+3 actually generate POINTS which are collinear. A vector equation, as the name implies, generates VECTORS, and these vectors are definitely NOT COLLINEAR.

How then can we say that an equation such as
r = (2,1,3) + t(1,2,4) is the "equation of a line"?

Also, why is it not possible to produce a scalar equation for a line in 3-D?

[pmb_phy]

Originally posted by Saad
Scalar equations such as y=2x+3 actually generate POINTS which are collinear. A vector equation, as the name implies, generates VECTORS, and these vectors are definitely NOT COLLINEAR.

How then can we say that an equation such as
r = (2,1,3) + t(1,2,4) is the "equation of a line"?

Also, why is it not possible to produce a scalar equation for a line in 3-D?

Consider the physical meaning of the vector r. That is called the position vector. It represents a spatial displacement from a point called the origin. The tip of the vector defines a point and it is that point we are refering to as the position.

Since r is the position vector which traces out a line, i.e. the tip of the vector traces out a line, the its called the equation of a line. Likewise the tip of the vector

\mathbf{r} = cos \theta \mathbf{i} + sin \theta \mathbf{j}

traces out a circle. Therefore one can say that this is the equation of a circle.

[matt grime]

It is possible to produce a set of scalar equations that generate a line in R^3

eg the line (1,2,3) + t(2,3,1) is also described as

(x- 1)/2 = (y-2)/3 = z-3


just as the original scalar equation you gave is expressible as a vector equation:

L = (0,3) +t(1,2)

[selfAdjoint]

It is possible to produce a set of scalar equations that generate a line in R^3

eg the line (1,2,3) + t(2,3,1) is also described as

(x- 1)/2 = (y-2)/3 = z-3


just as the original scalar equation you gave is expressible as a vector equation:

L = (0,3) +t(1,2)

This is called a parametric rep, with t as the parameter. For three dimension look up "direction cosines".