DIRECTOR VECTOR: EQUATION OF THE LINE, SOLVED EXERCISES - PHYSICAL - 2025

A director vector is understood to be one that defines the direction of a line, either in the plane or in space. Therefore, a vector parallel to the line can be considered as a directing vector of it.

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

Figure 1. Director vector of a line. (Own elaboration)

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

Equation of the line and director vector

Figure 2. Equation of the line and director vector. (Own elaboration)

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

PQ = t⋅ u

in the above expression t is a parameter that belongs to the real numbers.

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

(X-Xo, Y-Yo) = t⋅ (a, b)

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

X - Xo = a⋅ty Y - I = 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 director vector u = (a, b) are determined by assigning real values ​​to the variable parameter t:

{X = Xo + a⋅t; Y = I + b⋅t}

Example 1

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

u = (a, b) = (2, -1)

and as a known point of the line the point

P = (Xo, I) = (1, 5).

The parametric equation of the line is:

{X = 1 + 2⋅t; Y = 5 - 1⋅t; -∞

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

Figure 3. PQ = t u. (Own elaboration)

The line in vector form

Given a point P on the line and its director vector u, the equation of the line can be written in vector form:

OQ = OP + λ⋅ u

In the above equation, Q is any point but belonging to the line and λ is a real number.

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

In the three-dimensional case for a director 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 is:

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

Example 2

Consider again the line that has as a directing vector

u = (a, b) = (2, -1)

and as a known point of the line the point

P = (Xo, I) = (1, 5).

The vector equation of said line is:

(X, Y) = (1, 5) + λ⋅ (2, -1)

Continuous form of the line and the director vector

Starting from the parametric form, clearing and equating the parameter λ, we have:

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

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

Example 3

Consider the line that has as a directing vector

u = (a, b) = (2, -1)

and as a known point of the line the point

P = (Xo, I) = (1, 5). Find its symmetric shape.

The symmetric or continuous form of the line is:

(X - 1) / 2 = (Y - 5) / (- 1)

General form of the equation of the line

The general form of the line in the XY plane is known as the equation that has the following structure:

A⋅X + B⋅Y = C

The expression for the symmetric form can be rewritten to have the general form:

b⋅X - a⋅Y = b⋅Xo - a⋅Yo

comparing with the general shape of the line it is:

A = b, B = -a and C = b⋅Xo - a⋅Yo

Example 3

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

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

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

We begin by finding the dual vector w of the director vector u, defined as the vector obtained by exchanging the components of u and multiplying the second by -1:

w = (-1, -2)

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

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

(-1, -2) • (X, Y) = (-1, -2) • (1, 5)

-X-2Y = -1 -2⋅5 = -11

remaining finally:

X + 2Y = 11

Standard form of the equation of the line

It is known as the standard form of the line in the XY plane, one that has the following structure:

Y = m⋅X + d

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

Given the direction vector 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 / a and d = I - (b / a) Xo

Note that the slope m is the quotient between the y component of the director vector and the x component of it.

Example 4

Find the standard form of the line whose director vector is u = (2, -1)

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

m = -½ and 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 equation of the plane (Ω) clearance Y: Y = 1 -2X

Then we substitute in the equation of the plane (Π):

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

Then 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)

with which it is clear that the vector u = (1, -2, -3) is a directing 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 plane 5X + a Y + 4Z = 5

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

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

Solution 2

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

The vector u = (1, 3, -2) is a directing vector of the line.

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.

References

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