Two or more vectors are team lens if they have the same module, the same direction and the same sense, even if their point of origin is different. Remember that the characteristics of a vector are precisely: origin, module, direction and sense.
Vectors are represented by an oriented segment or arrow. Figure 1 shows the representation of several vectors in the plane, some of which are team-lensing according to the definition initially given..
At a first glance it is possible to see that the three green vectors have the same size, the same direction and the same sense. The same can be said about the two pink vectors and the four black vectors.
Many magnitudes of nature have a vector-like behavior, such is the case of speed, acceleration and force, to name just a few. Hence the importance of properly characterizing them.
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To distinguish vector quantities from scalar quantities, the bold typeface or an arrow above the letter is often used. When working with vectors by hand, on the notebook, it is necessary to distinguish them with the arrow and when using a printed medium, bold type is used.
Vectors can be denoted by indicating their point of departure or origin and their point of arrival. For example AB, BC, FROM Y EF of figure 1 are vectors, instead AB, BC, FROM Y EF are scalar quantities or numbers that indicate the magnitude, modulus or size of their respective vectors.
To indicate that two vectors are team-oriented, the symbol "∼ ". With this notation, in the figure we can point out the following vectors that are team-oriented to each other:
AB∼BC∼DE∼EF
They all have the same magnitude, direction and meaning. Therefore, they comply with the regulations indicated above..
Any of the vectors in the figure (for example AB) is representative of the set of all equipment-lens fixed vectors. This infinite set defines the class of free vectors or.
or = AB, BC, DE, EF, ...
An alternative notation is the following:
If the bold or the little arrow is not placed above the letter or, is that we want to refer to the vector module or.
Free vectors are not applied to any particular point.
For their part sliding vectors They are teamlength vectors to a given vector, but their point of application must be contained in the line of action of the given vector.
And the opposite vectors They are vectors that have the same magnitude and direction but opposite senses, although in English texts they are called de opposite directions since the direction also indicates the sense. Opposite vectors are not team.
Which other vectors of those shown in figure 1 are team-lenient to each other??
Apart from those already indicated in the previous section, it is observed from figure 1 that AD, BE Y EC They are also teamlength vectors:
AD ∼ BE ∼ CE
Any of them is representative of the class of free vectors v.
The vectors AE Y BF :
AE ∼ BF
Who are representatives of the class w.
Points A, B and C are on the Cartesian plane XY and their coordinates are:
A = (- 4.1), B = (- 1.4) and C = (- 4, -3)
Find the coordinates of a fourth point D such that the vectors AB Y CD be team-oriented.
So that CD be team-oriented AB must have the same module and the same address as AB .
The module AB squared is:
|AB| ^ 2 = (-1 - (-4)) ^ 2 + (4 -1) ^ 2 = 9 + 9 = 18
The coordinates of D are unknown so we can say: D = (x, y)
Then: |CD| ^ 2 = (x - (- 4)) ^ 2 + (y - (-3)) ^ 2
As |AB| = |CD| is one of the conditions for AB Y CD be team-oriented, you have:
(x + 4) ^ 2 + (y + 3) ^ 2 = 18
Since we have two unknowns, another equation is required, which can be obtained from the condition that AB Y CD are parallel and in the same sense.
The vector slope AB indicates your address:
Slope AB = (4 -1) / (- 1 - (-4)) = 3/3 = 1
Indicating that the vector AB 45º shape with the X axis.
The slope of CD is calculated in a similar way:
Slope CD = (y - (-3)) / (x - (- 4)) = (y + 3) / (x + 4)
Equating this result with the slope of AB we have the following equation:
y + 3 = x + 4
Which means that y = x + 1.
If this result is substituted in the equation for the equality of the modules, we have:
(x + 4) ^ 2 + (x + 1 + 3) ^ 2 = 18
Simplifying it remains:
2 (x + 4) ^ 2 = 18,
Which is equivalent to:
(x + 4) ^ 2 = 9
That is, x + 4 = 3 which implies that x = -1. So the coordinates of D are (-1, 0).
The components of the vector AB are (-1 - (- 4), 4 -1) = (3, 3)
and the vector CD are (-1 - (- 4)); 0 - (- 3)) = (3, 3)
Which means that the vectors are team-oriented. If two vectors have the same Cartesian components they have the same module and direction, therefore they are team-lensing.
The Free Vector or has magnitude 5 and direction 143.1301º.
Find its Cartesian components and determine the coordinates of points B and C knowing that the fixed vectors AB and CD are team-oriented to u. The coordinates of A are (0, 0) and the coordinates of point C are (-3,2).
The situation posed by the exercise can be represented by the following figure:
The Cartesian components of or They are
or = (5 * cos (143.1301º); 5 * sin (143.1301º))
Doing the calculations it remains:
or = (-4,3)
The coordinates of B are unknown so we will place B (x, y)
The coordinates of the vector AB are (x-0; y-0), but since u is team-lensing, the equality of components must be fulfilled, it follows therefore that the coordinates of B are (-4, 3).
Similarly the coordinates of the vector CD are (x - (- 3)); (and - 2) that it must be team-oriented u, lor leading to:
x + 3 = -4 and y -2 = 3
Then the coordinates of point D will be (-7, 5).
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