Vector Director: Line Equation, Solved Exercises

This is possible thanks to an axiom of Euclidean geometry that two points define a line. Then the oriented segment formed by these two points also defines an initial vector of that line.

Given a point P belonging to line (L) and given a direct vector u of that line, the line is completely determined.

Line vector equation and director

Given a point P of coordinates P: (Xo, I) and a vector or director of a line (L) , every point Q of coordinates Q: (X, Y) must fulfill that vector PQ is parallel to u. This last condition is guaranteed if PQ is proportional to u :

PQ = t⋅ u

In the previous expression, t is a parameter that belongs to real numbers.

If the Cartesian components of PQ and u are written, the above equation will be written as follows:

(X-Xo, Y-Yo) = t⋅ (a, b)

If the components of vector equality are matched, the following pair of equations is obtained:

X – Xo = a⋅t and Y – Yo = b⋅t

Parametric equation of the line

The X and Y coordinates of a point belonging to the line (L) that passes through a coordinate point (Xo, Yo) and is parallel to the main vector u = (a, b) are determined by assigning real values ​​to the variable parameter t :

{X = Xo + a⋅t; Y = Yo + b⋅t}

Example 1

To illustrate the meaning of the parametric equation of the line, we take as the main vector

u = (a, b) = (2, -1)

and as the known point of the line, the point

P = (Xo, Yo) = (1, 5) .

The parametric equation of the line is:

{X = 1 + 2⋅t; Y = 5 – 1⋅t; -∞<t<∞}

To illustrate the meaning of this equation, Figure 3 is shown, where the t parameter changes value and the Q coordinate point (X, Y) assumes different positions on the line.

The line in vector form

Given a point P of the line and its main vector, u can write the equation of the line in vector form:

OQ = OP + λ⋅ u

In the previous equation, Q is any point, but it belongs to the line and λ is a real number.

The line vector equation is applicable to any number of dimensions, even a hyperline can be defined.

In the three-dimensional case of an initial vector u = (a, b, c) and a point P = (Xo, Yo, Zo) , the coordinates of a generic point Q = (X, Y, Z) belonging to the line are: :

(X, Y, Z) = (Xo, Yo, Zo) + λ⋅ (a, b, c)

Example 2

Consider again the line with the main vector

u = (a, b) = (2, -1)

and as the known point of the line, the point

P = (Xo, Yo) = (1, 5) .

The vector equation of that line is:

(X, Y) = (1, 5) + λ⋅ (2, -1)

Line continuous shape and the director vector

From the parametric way, clearing and combining the λ parameter, you have:

(X-Xo) / a = (Y-Yo) / b = (Z-Zo) / c

This is the symmetric form of the straight line equation. Note that a , b and c are the components of the direction vector.

Example 3

Consider the line you have as the main vector

u = (a, b) = (2, -1)

and as the known point of the line, the point

P = (Xo, Yo) = (1, 5) . Find your symmetrical shape.

The symmetrical or continuous shape is of the line is:

(X – 1) / 2 = (Y – 5) / (- 1)

General form of the straight equation

It is known as the general shape of the line in the XY plane of the equation which has the following structure:

A⋅X + B⋅Y = C

The expression of the symmetrical shape can be rewritten so that it has the general shape:

b⋅X – a⋅Y = b⋅Xo – a⋅Yo

comparing with the general shape of the line is:

A = b, B = -a and C = b⋅Xo – a⋅Yo

Example 3

Find the general shape of the line whose main vector is u = (2, -1)

and that passes through the point P = (1, 5).

To find the general form we can use the formulas provided, however an alternative path will be chosen.

We start by finding the double vector w of the director vector u, defined as the vector obtained by swapping the components of u and multiplying by -1 the second:

w = (-1; -2)

the double vector w corresponds to a 90° clockwise rotation of the director vector v .

We scalarly multiply w by (X, Y) and by (Xo, Yo) and equal:

(-1, -2) • (X, Y) = (-1, -2) • (1, 5)

-X-2Y = -1 -2⋅5 = -11

finally remaining:

X + 2Y = 11

Standard form of the line equation

It is known as the standard shape of the line in the XY plane, which has the following structure:

Y = m⋅X + d

where m represents the slope and d the intercept with the Y axis.

Given the vector director u = (a, b), the slope m is b / a.

Y d is obtained by substituting X and Y for the known point Xo, I:

I = (b / a) Xo + d.

In short, m = b / ayd = I – (b / a) Xo

Note that the slope m is the quotient between the y component of the main vector and its x component .

Example 4

Find the default shape of the line whose main vector is u = (2, -1)

and that passes through the point P = (1, 5).

m = -½ and d d = 5 – (-½) 1 = 11/2

Y = (-1/2) X + 11/2

solved exercises

-Exercise 1

Find a director vector of the line (L) that is the intersection of the plane (Π): X – Y + Z = 3 and the plane (Ω): 2X + Y = 1.

Then write the continuous form of the equation of the line (L).

Solution

From the plane equation (Ω), Y offset: Y = 1-2X

Then we substitute in the plane equation (Π):

X – (1 – 2X) + Z = 3 ⇒ 3X + Z = 4 ⇒ Z = 4-3X

So we parameterize X, we choose the parameterization X = λ

This means that the line has a vector equation given by:

(X, Y, Z) = (λ, 1-2λ, 4-3λ)

which can be rewritten as:

(X, Y, Z) = (0, 1, 4) + λ (1, -2, -3)

then it is clear that the vector u = (1, -2, -3) is a direct vector of the line (L).

The continuous form of the line (L) is:

(X – 0) / 1 = (Y – 1) / (- 2) = (Z – 4) / (- 3)

-Exercise 2

Given the plan 5X + in Y + 4Z = 5

and the line whose equation is X / 1 = (Y-2) / 3 = (Z -2) / (-2)

Determine the value of a so that the plane and the line are parallel.

Solution 2

The vector n = (5, a, 4) is a normal vector for the plane.

The vector u = (1, 3, -2) is a straight line vector.

If the line is parallel to the plane, then n • v = 0.

(5, a , 4) • (1, 3, -2) = 5 +3 a -8 = 0 ⇒ a = 1.