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 $\rv{-2,-6}+\lambda\cdot\rv{1,-4}$ .

The position vector $\rv{-2,-6}$ is drawn in blue. The directional vector $\rv{1,-4}$ and the line $\ell$ are drawn in black..

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