What Is V1, the Vector From Point O to Point E?

2 Vectors

2.2 Coordinate Systems and Components of a Vector

Learning Objectives

By the end of this section, you will be able to:

Describe vectors in ii and iii dimensions in terms of their components, using unit of measurement vectors along the axes.
Distinguish betwixt the vector components of a vector and the scalar components of a vector.
Explicate how the magnitude of a vector is defined in terms of the components of a vector.
Identify the management angle of a vector in a aeroplane.
Explain the connectedness between polar coordinates and Cartesian coordinates in a plane.

Vectors are usually described in terms of their components in a coordinate system . Even in everyday life nosotros naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For case, if y'all ask someone for directions to a particular location, yous will more likely exist told to go xl km east and thirty km n than l km in the management

$37\text{°}$

north of east.

In a rectangular (Cartesian) xy-coordinate system in a airplane, a point in a aeroplane is described by a pair of coordinates (ten, y). In a similar manner, a vector

$\overset{\to }{A}$

in a aeroplane is described past a pair of its vector coordinates. The ten-coordinate of vector

$\overset{\to }{A}$

is chosen its 10-component and the y-coordinate of vector

$\overset{\to }{A}$

is called its y-component. The vector ten-component is a vector denoted by

${\overset{\to }{A}}_{x}$

. The vector y-component is a vector denoted by

${\overset{\to }{A}}_{y}$

. In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x– and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane tin can be expressed as the vector sum of its vector components:

$\overset{\to }{A}={\overset{\to }{A}}_{x}+{\overset{\to }{A}}_{y}.$

As illustrated in (Figure), vector

$\overset{\to }{A}$

is the diagonal of the rectangle where the x-component

${\overset{\to }{A}}_{x}$

is the side parallel to the x-axis and the y-component

${\overset{\to }{A}}_{y}$

is the side parallel to the y-axis. Vector component

${\overset{\to }{A}}_{x}$

is orthogonal to vector component

${\overset{\to }{A}}_{y}$

Vector A is shown in the x y coordinate system and extends from point b at A's tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b. — **Figure 2.16** Vector
$\overset{\to }{A}$

in a aeroplane in the Cartesian coordinate system is the vector sum of its vector x- and y-components. The x-vector component

${\overset{\to }{A}}_{x}$

is the orthogonal projection of vector

$\overset{\to }{A}$

onto the x-axis. The y-vector component

${\overset{\to }{A}}_{y}$

is the orthogonal projection of vector

$\overset{\to }{A}$

onto the y-axis. The numbers

${A}_{x}$

and

${A}_{y}$

that multiply the unit of measurement vectors are the scalar components of the vector.

It is customary to denote the positive direction on the 10-axis by the unit vector

$\hat{i}$

and the positive direction on the y-axis by the unit vector

$\hat{j}$

. Unit vectors of the axes,

$\hat{i}$

and

$\hat{j}$

, define two orthogonal directions in the plane. As shown in (Figure), the 10– and y– components of a vector can now be written in terms of the unit vectors of the axes:

$\{\begin{array}{c}{\overset{\to }{A}}_{x}={A}_{x}\hat{i}\\ {\overset{\to }{A}}_{y}={A}_{y}\hat{j}.\end{array}$

The vectors

${\overset{\to }{A}}_{x}$

and

${\overset{\to }{A}}_{y}$

defined by (Figure) are the vector components of vector

$\overset{\to }{A}$

. The numbers

${A}_{x}$

and

${A}_{y}$

that define the vector components in (Figure) are the scalar components of vector

$\overset{\to }{A}$

. Combining (Figure) with (Figure), we obtain the component class of a vector:

$\overset{\to }{A}={A}_{x}\hat{i}+{A}_{y}\hat{j}.$

If we know the coordinates

$b({x}_{b},{y}_{b})$

of the origin indicate of a vector (where b stands for "beginning") and the coordinates

$e({x}_{e},{y}_{e})$

of the finish signal of a vector (where e stands for "end"), we can obtain the scalar components of a vector but by subtracting the origin point coordinates from the end point coordinates:

$\{\begin{array}{c}{A}_{x}={x}_{e}-{x}_{b}\\ {A}_{y}={y}_{e}-{y}_{b}.\end{array}$

Example

Displacement of a Mouse Pointer

A mouse pointer on the brandish monitor of a reckoner at its initial position is at point (half-dozen.0 cm, 1.six cm) with respect to the lower left-side corner. If you move the pointer to an icon located at betoken (ii.0 cm, 4.5 cm), what is the deportation vector of the pointer?

Strategy

The origin of the xy-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit of measurement vector

$\hat{i}$

on the ten-axis points horizontally to the correct and the unit vector

$\hat{j}$

on the y-axis points vertically upward. The origin of the deportation vector is located at betoken b(6.0, ane.6) and the end of the deportation vector is located at point due east(ii.0, four.5). Substitute the coordinates of these points into (Figure) to observe the scalar components

${D}_{x}$

and

${D}_{y}$

of the displacement vector

$\overset{\to }{D}$

. Finally, substitute the coordinates into (Figure) to write the displacement vector in the vector component form.

Solution

[reveal-reply q="907005″]Bear witness Answer[/reveal-reply]
[hidden-answer a="907005″]We identify

${x}_{b}=6.0$

${x}_{e}=2.0$

${y}_{b}=1.6$

, and

${y}_{e}=4.5$

, where the concrete unit of measurement is i cm. The scalar 10- and y-components of the displacement vector are

$\begin{array}{cc}\hfill {D}_{x}& ={x}_{e}-{x}_{b}=(2.0-6.0)\text{cm}=-4.0\,\text{cm},\hfill \\ \hfill {D}_{y}& ={y}_{e}-{y}_{b}=(4.5-1.6)\text{cm}=+2.9\,\text{cm}.\hfill \end{array}$

The vector component form of the displacement vector is

$\overset{\to }{D}={D}_{x}\hat{i}+{D}_{y}\hat{j}=(-4.0\,\text{cm})\hat{i}+(2.9\,\text{cm})\hat{j}=(-4.0\hat{i}+2.9\hat{j})\text{cm}.$

This solution is shown in (Figure ii.17).

Vector D extends from coordinates 6.0, 1.6 to coordinates 2.0, 4.5. Vector D equals vector D sub x plus vector D sub y. D sub x equals minus 4.0 I hat, and extends from x=6.0 to x =2.0. The magnitude D sub x equals 2.0-6.0 = -4.0. D sub y equals plus 2.9 j hat, and extends from y=1.6 to y=4.5. The magnitude D sub y equals 4.5 − 1.6. — **Figure 2.17** The graph of the displacement vector. The vector points from the origin point at b to the end signal at e.

[/hidden-answer]

Significance

Observe that the physical unit—here, ane cm—tin be placed either with each component immediately before the unit vector or globally for both components, every bit in (Figure). Often, the latter fashion is more user-friendly considering it is simpler.

The vector 10-component

${\overset{\to }{D}}_{x}=-4.0\hat{i}=4.0(\text{−}\hat{i})$

of the deportation vector has the magnitude

$|{\overset{\to }{D}}_{x}|=|-4.0||\hat{i}|=4.0$

because the magnitude of the unit vector is

$|\hat{i}|=1$

. Notice, too, that the management of the x-component is

$\text{−}\hat{i}$

, which is antiparallel to the direction of the +10-axis; hence, the x-component vector

${\overset{\to }{D}}_{x}$

points to the left, every bit shown in (Figure). The scalar 10-component of vector

$\overset{\to }{D}$

${D}_{x}=-4.0$

Similarly, the vector y-component

${\overset{\to }{D}}_{y}=+2.9\hat{j}$

of the displacement vector has magnitude

$|{\overset{\to }{D}}_{y}|=|2.9||\hat{j}|=\,2.9$

because the magnitude of the unit vector is

$|\hat{j}|=1$

. The management of the y-component is

$+\hat{j}$

, which is parallel to the management of the +y-centrality. Therefore, the y-component vector

${\overset{\to }{D}}_{y}$

points upward, every bit seen in (Effigy). The scalar y-component of vector

$\overset{\to }{D}$

${D}_{y}=+2.9$

. The displacement vector

$\overset{\to }{D}$

is the resultant of its 2 vector components.

The vector component form of the displacement vector (Figure) tells us that the mouse pointer has been moved on the monitor four.0 cm to the left and two.9 cm upwards from its initial position.

Bank check Your Understanding

A blue fly lands on a sheet of graph newspaper at a signal located x.0 cm to the correct of its left edge and viii.0 cm higher up its bottom edge and walks slowly to a point located 5.0 cm from the left border and five.0 cm from the bottom edge. Choose the rectangular coordinate organisation with the origin at the lower left-side corner of the newspaper and discover the displacement vector of the fly. Illustrate your solution by graphing.

[reveal-answer q="fs-id1167132407980″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1167132407980″]

$\overset{\to }{D}=(-5.0\hat{i}-3.0\hat{j})\text{cm}$

; the fly moved 5.0 cm to the left and iii.0 cm downwardly from its landing site.
[/hidden-answer]

When we know the scalar components

${A}_{x}$

and

${A}_{y}$

of a vector

$\overset{\to }{A}$

, we can observe its magnitude A and its direction angle

${\theta }_{A}$

. The direction bending—or direction, for short—is the angle the vector forms with the positive management on the ten-centrality. The angle

${\theta }_{A}$

is measured in the counterclockwise management from the +x-axis to the vector ((Effigy)). Considering the lengths A,

${A}_{x}$

, and

${A}_{y}$

form a right triangle, they are related by the Pythagorean theorem:

${A}^{2}={A}_{x}^{2}+{A}_{y}^{2}\enspace⇔\enspace{A}=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}.$

This equation works even if the scalar components of a vector are negative. The direction angle

${\theta }_{A}$

of a vector is defined via the tangent function of angle

${\theta }_{A}$

in the triangle shown in (Figure):

$\text{tan}\,{\theta }_{A}=\frac{{A}_{y}}{{A}_{x}}\enspace⇒\enspace{\theta }_{A}={\text{tan}}^{-1}(\frac{{A}_{y}}{{A}_{x}}).$

Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A. — **Figure ii.xviii** For vector
$\overset{\to }{A}$

, its magnitude A and its direction angle

${\theta }_{A}$

are related to the magnitudes of its scalar components considering A,

${A}_{x}$

, and

${A}_{y}$

form a right triangle.

When the vector lies either in the first quadrant or in the fourth quadrant, where component

${A}_{x}$

is positive ((Figure)), the bending

$\theta$

in (Figure) is identical to the direction angle

${\theta }_{A}$

. For vectors in the quaternary quadrant, angle

$\theta$

is negative, which means that for these vectors, direction bending

${\theta }_{A}$

is measured clockwise from the positive 10-axis. Similarly, for vectors in the second quadrant, angle

$\theta$

is negative. When the vector lies in either the second or tertiary quadrant, where component

${A}_{x}$

is negative, the direction bending is

${\theta }_{A}=\theta +180\text{°}$

((Effigy)).

Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees. — **Figure two.19** Scalar components of a vector may exist positive or negative. Vectors in the get-go quadrant (I) accept both scalar components positive and vectors in the third quadrant accept both scalar components negative. For vectors in quadrants II and III, the direction bending of a vector is
${\theta }_{A}=\theta +180\text{°}$

.

\[{\theta }_{A}=\theta +180\text{°}\] — **Figure two.19** Scalar components of a vector may exist positive or negative. Vectors in the get-go quadrant (I) accept both scalar components positive and vectors in the third quadrant accept both scalar components negative. For vectors in quadrants II and III, the direction bending of a vector is
${\theta }_{A}=\theta +180\text{°}$

.