# vector equation of a line

• Jul 2nd 2010, 04:02 AM
Stuck Man
vector equation of a line
A vector equation of a line in symmetric form is either written with =t at the right end usually or omitted. Is this a matter of preference?

Also are the terms symmetric form and Cartesian form referring to the same thing? Is Cartesian form the more professional term?
• Jul 2nd 2010, 08:19 AM
Soroban
Hello, Stuck Man!

The line through the point $\displaystyle (x_1,y_1,z_1)$ with direction $\displaystyle \vec v = \langle a,b,c\rangle$
can be written in two forms:

. . Symmetric form: .$\displaystyle \dfrac{x-x_1}{a} \:=\:\dfrac{y-y_1}{b} \:=\:\dfrac{z-z_1}{c}$

. . Cartesian form: .$\displaystyle \begin{Bmatrix}x &=& x_1 + at \\ y &=& y_1 + bt \\ z &= & z_1+ct \end{Bmatrix}$

Quote:

A vector equation of a line in symmetric form is either written
with = t at the right end usually or omitted.
Is this a matter of preference?

The symmetric form is derived from the Cartesian form.

$\displaystyle \text{Solve the Cartesian equations for }t\!:$

. . $\displaystyle \begin{array}{ccccc}x \:=\: x_1 + at & \Rightarrow & t \:=\:\frac{x-x_1}{a} & [1]\\ \\[-4mm] y \:=\:y_1 + bt & \Rightarrow & t \:=\:\frac{y-y_1}{b} & [2] \\ \\[-4mm] z \:=\:z_1+ ct & \Rightarrow & t \:=\:\frac{z-z_1}{c} & [3]\end{array}$

Equate [1], [2], [3]: .$\displaystyle \dfrac{x-x_1}{a} \;=\;\dfrac{y-y_1}{b} \;=\;\dfrac{z-z_1}{c}$

The $\displaystyle t$ is usually omitted.

Quote:

Also are the terms symmetric form and Cartesian form referring to the same thing?
Is Cartesian form the more professional term?

Both forms refer to the same line.

The Cartesian form is more convenient for algebraic work.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

An amusing note . . . well, I think it's funny.

Given the point $\displaystyle (3,1,4)$ and direction $\displaystyle \vec v = \langle5,2,0\rangle$

. . the symmetric equation is: .$\displaystyle \dfrac{x-3}{5} \:=\:\dfrac{y-1}{2} \:=\:\dfrac{z-4}{0}$

This may be the only instance where "division by zero" is defined.

• Jul 2nd 2010, 09:01 AM
Stuck Man
I've seen a couple of books where the Cartesian form is identical to the symmetrical form you have given an example of. What you have called the Cartesian form is the parametric form.

I've noticed that about dividing by zero.
• Jul 2nd 2010, 09:21 AM
Stuck Man
The symmetric form of the line through $\displaystyle (3,1,4)$ with direction vector $\displaystyle \left\langle {5,2,0} \right\rangle$ is $\displaystyle \frac{{x - 3}}{5} = \frac{{y - 1}}{2};~z = 4$