# Vector Director: Line Equation, Solved Exercises

A **director vector is** defined as one that defines the direction of a line, in plane or in space. Therefore, a vector parallel to the line can be thought of as a direction vector.

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

**is proportional to**

*PQ***:**

*u**PQ** = t⋅ **u*

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

If the Cartesian components of ** PQ** and

**are written, the above equation will be written as follows:**

*u**(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.