Straight lines and planes

Vector calculus in plane and space: Straight Lines and Planes

Straight lines and planes

From now on we assume that we have chosen an origin. The origin itself corresponds to the zero vector #\vec{0}#.

Parametric representation of a line

Let #\vec{v}# be a vector that is not equal to #\vec{0}#. The multiples #\vec{x}= \lambda \cdot\vec{v}# of #\vec{v}# run over the points/vectors of a straight line #\ell# through the origin. We call this line the line spanned by #\vec{v}#.

If #\vec{a}# is a second vector, then the point #\vec{x} = \vec{a}+\lambda\cdot \vec{v}#, for , then #\lambda# varying over the real numbers, runs over the points of the line #m# through #\vec{a}# parallel to #\ell#. We call \[\ell : \,\vec{x} = \lambda\cdot\vec{v}\phantom{x} \hbox{ and }\phantom{x}m :\, \vec{x} = \vec{a} + \lambda \cdot\vec{v}\] a parametric or vector representation of the line #\ell# and #m#, resectively.

The vector #\vec{v}# is a so-called direction vector in both cases. The vector #\vec{a}# is a particular vector of the parametric representation of the line #m#. We call #\lambda# the parameter.

The line #\ell# is the special case of the general line #m# wherein #\vec{a}=\vec{0}#. The vector #\vec{0}# is a support vector of the line #\ell#.

The position and direction vector of a line are not unique:

The vector #\vec{p} + \vec{v}# is located on the line #\ell# with parametric representation #\vec{x} = \vec{p} + \lambda \vec{v}#. Try entering #\lambda =1#. The vector #\vec{p}+\vec{v}# might as well act as a particular vector. The parametric representation \[\vec{x} = \vec{p}+\vec{v} + \mu \vec{v}\] is an equal parametric representation of the line #\ell#. For varying #\mu#, #\vec{p}+\vec{v} + \mu \vec{v}# in fact crosses the same vectors as #\vec{p} + \lambda \cdot\vec{v}# for varying #\lambda# (check). In fact, any vector can act as particular vector on #\ell#.
In addition to #\vec{v}#,#2\vec{v}#, #-3\vec{v}#, #\pi \cdot \vec{v}# are direction vectors of #\ell#. The vectors of #\vec{p} + \mu\cdot (2\vec{v})# exactly cross the points of #\ell# at varying #\mu#.

Instead of particular vector, we will speak of position point. This is, after all, a point in the plane, which is the end point of the representative of the particular vector which has its starting point in the origin.

To summarize: in order for a parametric representation of a line, a particular vector and a direction vector are required.

Criteria for two vectors on a line through the origin

Two vectors can only be on a line through the origin if (at least) one of the two is a scalar multiple of the other.

Let #\vec{v}# and #\vec{w}# be two vectors. If they are the same, then they are located on a line through the origin, and one is a scalar multiple (with scalar #1# ) of the other. For the remaining proof, we can therefore assume that the two vectors are not equal.

Let #\ell# be the line through #\vec{v}# and #\vec{w}#. This line has parametric representation \[\lambda\cdot\vec{v}+\left(1-\lambda\right)\cdot\vec{w}\tiny.\]

First, suppose that #\vec{w}=\mu\cdot\vec{v}#, so #\vec{w}# is a scalar multiple of #\vec{v}# with scalar #\mu#. Because #\vec{v}# and #\vec{w}# are different, #\mu\ne1# applies so we can divide by #\mu-1#. Therefore, we can choose #\lambda=\frac{\mu}{\mu-1}# for: \[\begin{array}{rcl}\frac{\mu}{\mu-1}\cdot\vec{v}+\left(1-\frac{\mu}{\mu-1}\right)\cdot\vec{w}&=&\frac{\mu}{\mu-1}\cdot\vec{v}-\frac{1}{\mu-1}\cdot\vec{w}\\&=&\frac{\mu}{\mu-1}\cdot \vec{v}-\frac{\mu}{\mu-1}\cdot\vec{v}\\&=&\vec{0}\tiny.\end{array}\] The origin can be written in the parametric representation above, of #\ell#; i.e., #\vec{0}# is located on #\ell#.

We now determine the reverse. We assume that #\vec{0}# is located on #\ell#, and deduce that one of the two vectors #\vec{v}#, #\vec{w}# is a scalar multiple of the other. The parametric equation above gives us the scalar #\lambda#, so \[\vec{0}=\lambda\cdot\vec{v}+\left(1-\lambda\right)\cdot\vec{w}\tiny.\] If #\lambda\ne0#, what follows after rewriting #\vec{v} = \frac{\lambda-1}{\lambda}\cdot \vec{w}#. If #\lambda\ne1#, what follows after rewriting is #\vec{w} = \frac{\lambda}{\lambda-1}\cdot \vec{v}#. Because (at least) one of the two, #\lambda\ne0# or #\lambda\ne1#, is true, we see that one of the two vectors #\vec{v}#, #\vec{w}# is a scalar multiple of the other.

This proves the theorem.

Draw the line #\ell# with parametric representation #\rv{5,1}+\lambda\cdot\rv{3,-6}#.

The position vector #\rv{5,1}# is drawn in blue. The directional vector #\rv{3,-6}# and the line #\ell# are drawn in black..

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