Symmetric Equation of a Line

There are several ways to denote a distinct line in R3. Symmetric equation is one such way of representing distinct lines. Initially to deal with symmetric equation familiarity with parametric equation is necessary. Obviously this notation is considered compact too. Getting into the concept of symmetric equation, it sets each component of the line equal to a common parameter. Following this all the components are set equal to one another. Knowledge in vector equation of line and parametric equation would help us in solving tutorials.

More about Symmetric Matrix Equation of a Line:

The parametric equation of the line is stated as follows:

`(x-a_x)/(m_1)=(y-a_y)/(m_2)=(z-a_z)/(m_3)`

To arrive at this equation, each expression in the parametric equation of the line should be treated equal. For example

`x: a_x + (t)(m_1)`

It should be taken as

`x=a_x + (t)(m_1)`

The expression should be solved for each value of t and ultimately the resultant expression would be equal to the constant t and therefore would be equal to each other. Interestingly this expression of line violates one of the common doctrines of mathematics which limits us to use only one equality sign per expression. It is this factor that helps us to use this mode of representation in fewer or more dimensions. In case we are to use it in R2 form then the z expression would be ignored thus containing only one equality sign per expression. Similarly if we are expected to express a line that is parallel to a co ordinate plane then the term of the axis to which it is parallel to is also ignored.
Example Problems – Symmetric Equation of a Line:

Example 1 – Symmetric equation of a line:

Find the symmetric equation of the line through point (6,-7,20) & perpendicular to the plane 2x+3y-6z-8=0.

Solution:

Let us consider a normal vector n

2x + 3y - 6z - 8 = 0 is n = <2 -6="-6" 3="3">

The vector n is also the directional vector of the line thru point P(6, -7, 20) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <6 -7="-7" 20="20"> + t<2 -6="-6" 3="3">

L = <6 -7="-7" -="-" 20="20" 2="2" 3t="3t" 6t="6t" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 6 + 2t

y = -7 + 3t

z = 20 - 6t

Symmetric form of equation.

`t` =` (x - 6)/2` = `(y + 7)/3` = `(z - 20)/-6`


Example 2 – Symmetric equation of a line:

Find the symmetric equation of the line through point (2, 3, 4) & perpendicular to the plane x+2y-3z-4=0.

Solution:

Let us consider a normal vector n

x+2y-3z-4=0 is n = <1 -3="-3" 2="2">

The vector n is also the directional vector of the line thru point P(2, 3, 4) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <2 3="3" 4="4"> + t<1 -3="-3" 2="2">

L = <2 -="-" 2t="2t" 3="3" 3t="3t" 4="4" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 2 + t

y = 3 + 2t

z = 4 - 3t

Symmetric form of equation.

`t` = `(x - 2)/1` = `(y - 3)/2` = `(z - 4)/-3`