It is understood by director vector 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 the same.
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.
Given a point P belonging to the line (L) and given a director vector or of that line, the line is completely determined.
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Given a point P of coordinates Q: (Xo, I) and a vector or director of a straight (L), all point Q of coordinates Q: (X, Y) must satisfy that the vector PQ be parallel to u. This last condition is guaranteed if PQ is proportional to or:
PQ = t⋅or
in the previous expression t is a parameter that belongs to the real numbers.
If the Cartesian components of PQ and of or The above equation is written as follows:
(X-Xo, Y-Yo) = t⋅ (a, b)
If the components of vector equality are equalized, we have the following pair of equations:
X - Xo = a⋅t Y Y - I = b⋅t
The coordinates X and Y of a point on the line (L) passing through a coordinate point (Xo, I) and it is parallel to director vector or= (a, b) are determined by assigning real values to the variable parameter t:
X = Xo + a⋅t; Y = I + b⋅t
To illustrate the meaning of the parametric equation of the line, we take as the directing vector
or = (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) take different positions on the straight.
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⋅or
In the above equation Q is any point but belonging to the line and λ 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 or= (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 AND Z) = (Xo, Yo, Zo) + λ⋅ (a, b, c)
Consider again the line that has as a directing vector
or = (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)
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. I feel that to, b Y c are the components of the director vector.
Consider the line that has as a directing vector
or = (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)
The equation that has the following structure is known as the general form of the line in the XY plane:
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
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 multiply scalarly w with (X, Y) and with (Xo, I) and we match:
(-1, -2) • (X, Y) = (-1, -2) • (1, 5)
-X-2Y = -1 -2⋅5 = -11
remaining finally:
X + 2Y = 11
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 component Y of the director vector and the component x of the same.
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
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).
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 or = (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)
Given the 5X plane + to Y + 4Z = 5
and the line whose equation is X / 1 = (Y-2) / 3 = (Z -2) / (- 2)
Determine the value of to so that the plane and the line are parallel.
The vector n = (5, a, 4) is a vector normal to the plane.
The vector or = (1, 3, -2) is a director vector of the line.
If the line is parallel to the plane, then n • v = 0.
(5, to, 4)•(1, 3, -2) = 5 +3to -8 = 0 ⇒ to= 1.
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