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10.8: Vectors - Mathematics


Sources: JayAbramson(OpenStax)-Bookshelves/Algebra/Book:_Algebra_and_Trigonometry_/10:_Further_Applications_of_Trigonometry/10.9:_Vectors

" Vectors" by Jay Abramson, LibreTexts is licensed under CC BY . (2021, January 2). e

Learning Objectives

  • View vectors geometrically.
  • Find magnitude and direction.
  • Perform vector addition and scalar multiplication.
  • Find the component form of a vector.
  • Find the unit vector in the direction of (v).
  • Perform operations with vectors in terms of (i) and (j).
  • Find the dot product of two vectors.

An airplane is flying at an airspeed of (200) miles per hour headed on a SE bearing of (140°). A north wind (from north to south) is blowing at (16.2) miles per hour, as shown in Figure (PageIndex{1}). What are the ground speed and actual bearing of the plane?

Ground speed refers to the speed of a plane relative to the ground. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s ground speed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors.

A Geometric View of Vectors

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:

  • Lower case, boldfaced type, with or without an arrow on top such as (v), (u), (w), (overrightarrow{v}), (overrightarrow{u}), (overrightarrow{w}).
  • Given initial point (P) and terminal point (Q), a vector can be represented as (overrightarrow{PQ}). The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
  • Given an initial point of ((0,0)) and terminal point ((a,b)), a vector may be represented as (⟨a,b⟩).

This last symbol (⟨a,b⟩) has special significance. It is called the standard position. The position vector has an initial point ((0,0)) and a terminal point (⟨a,b⟩). To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector (overrightarrow{CD}) is (C(x_1,y_1)) and the terminal point is (D(x_2,y_2)), then the position vector is found by calculating

[egin{align*} overrightarrow{AB} &= ⟨x_2−x_1,y_2−y_1⟩ [4pt] &= ⟨a,b⟩ end{align*}]

In Figure (PageIndex{2}), we see the original vector (overrightarrow{CD}) and the position vector (overrightarrow{AB}).

PROPERTIES OF VECTORS

A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at ((0,0)) and is identified by its terminal point (⟨a,b⟩).

Example (PageIndex{1A}): Find the Position Vector

Consider the vector whose initial point is (P(2,3)) and terminal point is (Q(6,4)). Find the position vector.

Solution

The position vector is found by subtracting one (x)-coordinate from the other (x)-coordinate, and one (y)-coordinate from the other (y)-coordinate. Thus

[egin{align*} v &= ⟨6−2,4−3⟩ [4pt] &=⟨4,1⟩ end{align*}]

The position vector begins at ((0,0)) and terminates at ((4,1)). The graphs of both vectors are shown in Figure (PageIndex{3}).

We see that the position vector is (⟨4,1⟩).

Example (PageIndex{1B}): Drawing a Vector with the Given Criteria and Its Equivalent Position Vector

Find the position vector given that vector (v) has an initial point at ((−3,2)) and a terminal point at ((4,5)), then graph both vectors in the same plane.

Solution

The position vector is found using the following calculation:

[egin{align*} v &= ⟨4−(−3),5−2⟩ [4pt] &= ⟨7,3⟩ end{align*}]

Thus, the position vector begins at ((0,0)) and terminates at ((7,3)). See Figure (PageIndex{4}).

Exercise (PageIndex{1})

Draw a vector (vec{v}) that connects from the origin to the point ((3,5)).

Answer

Finding Magnitude and Direction

To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.

MAGNITUDE AND DIRECTION OF A VECTOR

Given a position vector (vec{v}=⟨a,b⟩),the magnitude is found by (| v |=sqrt{a^2+b^2}).The direction is equal to the angle formed with the (x)-axis, or with the (y)-axis, depending on the application. For a position vector, the direction is found by ( an heta=left(dfrac{b}{a} ight)⇒ heta={ an}^{−1}left(dfrac{b}{a} ight)), as illustrated in Figure (PageIndex{6}).

Two vectors (vec{v}) and (vec{u}) are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.

Example (PageIndex{2A}): Finding the Magnitude and Direction of a Vector

Find the magnitude and direction of the vector with initial point (P(−8,1)) and terminal point (Q(−2,−5)).Draw the vector.

Solution

First, find the position vector.

[egin{align*} u &= ⟨−2,−(−8),−5−1⟩ [4pt] &= ⟨6,−6⟩ end{align*}]

We use the Pythagorean Theorem to find the magnitude.

[egin{align*} |u| &= sqrt{{(6)}^2+{(−6)}^2} [4pt] &= sqrt{72} [4pt] &=sqrt{62} end{align*}]

The direction is given as

[egin{align*} an heta & =dfrac{−6}{6}=−1 ightarrow heta={ an}^{−1}(−1) [4pt] &= −45° end{align*}]

However, the angle terminates in the fourth quadrant, so we add (360°) to obtain a positive angle. Thus, (−45°+360°=315°). See Figure (PageIndex{7}).

Example (PageIndex{2B}): Showing That Two Vectors Are Equal

Show that vector (vec{v}) with initial point at ((5,−3)) and terminal point at ((−1,2)) is equal to vector (vec{u}) with initial point at ((−1,−3)) and terminal point at ((−7,2)). Draw the position vector on the same grid as (vec{v}) and (vec{u}). Next, find the magnitude and direction of each vector.

Solution

As shown in Figure (PageIndex{8}), draw the vector (vec{v}) starting at initial ((5,−3)) and terminal point ((−1,2)). Draw the vector (vec{u}) with initial point ((−1,−3)) and terminal point ((−7,2)). Find the standard position for each.

Next, find and sketch the position vector for (vec{v}) and (vec{u}). We have

[egin{align*} v &= ⟨−1−5,2−(−3)⟩ [4pt] &= ⟨−6,5⟩ [4pt] u&= ⟨−7−(−1),2−(−3)⟩ [4pt] & =⟨−6,5⟩ end{align*}]

Since the position vectors are the same, (vec{v}) and (vec{u}) are the same.

An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.

[egin{align*} |v| &= sqrt{{(−1−5)}^2+{(2−(−3))}^2} [4pt] &= sqrt{{(−6)}^2+{(5)}^2} [4pt] &= sqrt{36+25} [4pt] &= sqrt{61} [4pt] |u| &= sqrt{{(−7−(−1))}^2+{(2−(−3))}^2} [4pt] &=sqrt{{(−6)}^2+{(5)}^2} [4pt] &= sqrt{36+25} [4pt] &= sqrt{61} end{align*}]

As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives

[egin{align*} an heta &= −dfrac{5}{6}⇒ heta={ an}^{−1}left(−dfrac{5}{6} ight) [4pt] & = −39.8° end{align*}]

However, we can see that the position vector terminates in the second quadrant, so we add (180°). Thus, the direction is (−39.8°+180°=140.2°).

Performing Vector Addition and Scalar Multiplication

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector (u=⟨x,y⟩) as an arrow or directed line segment from the origin to the point ((x,y)), vectors can be situated anywhere in the plane. The sum of two vectors (vec{u}) and (vec{v}), or vector addition, produces a third vector (overrightarrow{u+ v}), the resultant vector.

To find (overrightarrow{u + v}), we first draw the vector (vec{u}), and from the terminal end of (vec{u}), we drawn the vector (vec{v}). In other words, we have the initial point of (vec{v}) meet the terminal end of (vec{u}). This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum (overrightarrow{u + v}) is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of (vec{u}) to the end of (vec{v}) in a straight path, as shown in Figure (PageIndex{9}).

Vector subtraction is similar to vector addition. To find (overrightarrow{u − v}), view it as (overrightarrow{u + (−v)}). Adding (overrightarrow{−v}) is reversing direction of (vec{v}) and adding it to the end of (vec{u}). The new vector begins at the start of (vec{u}) and stops at the end point of (overrightarrow{−v}). See Figure (PageIndex{10}) for a visual that compares vector addition and vector subtraction using parallelograms.

Example (PageIndex{3}): Adding and Subtracting Vectors

Given (u=⟨3,−2⟩) and (v=⟨−1,4⟩), find two new vectors (overrightarrow{u + v}), and (overrightarrow{u − v}).

Solution

To find the sum of two vectors, we add the components. Thus,

[ egin{align*} u+v &= ⟨3,−2⟩+⟨−1,4⟩ [4pt] &= ⟨3+(−1),−2+4⟩ [4pt] &=⟨2,2⟩ end{align*}]

See Figure (PageIndex{11a}).

To find the difference of two vectors, add the negative components of (vec{v}) to (vec{u}). Thus,

[egin{align*}u+(−v) &=⟨3,−2⟩+⟨1,−4⟩ [4pt] &= ⟨3+1,−2+(−4)⟩ [4pt] &= ⟨4,−6⟩ end{align*}]

See Figure (PageIndex{11b}).

Multiplying By a Scalar

While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.

SCALAR MULTIPLICATION

Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply (v=⟨a,b⟩) by (k), we have

(kv=⟨ka,kb⟩)

Only the magnitude changes, unless (k) is negative, and then the vector reverses direction.

Example (PageIndex{4}): Performing Scalar Multiplication

Given vector (vec{v}=⟨3,1⟩), find (3vec{v}), (dfrac{1}{2}), and (vec{−v}).

Solution

See Figure (PageIndex{12}) for a geometric interpretation. If (vec{v}=⟨3,1⟩), then

[egin{align*} 3v &= ⟨3⋅3,3⋅1⟩ [4pt] &= ⟨9,3⟩ [4pt] dfrac{1}{2}v &= ⟨dfrac{1}{2}⋅3,dfrac{1}{2}⋅1⟩ [4pt] &=⟨dfrac{3}{2},dfrac{1}{2}⟩ [4pt] −v &=⟨−3,−1⟩ end{align*}]

Analysis

Notice that the vector (3vec{v}) is three times the length of (vec{v}), (dfrac{1}{2}vec{v}) is half the length of (vec{v}), and (overrightarrow{–v}) is the same length of (vec{v}), but in the opposite direction.

Exercise (PageIndex{2})

Find the scalar multiple (3u) given (vec{u}=⟨5,4⟩).

Answer

(3u=⟨15,12⟩)

Example (PageIndex{5})

Given (vec{u}=⟨3,−2⟩) and (vec{v}=⟨−1,4⟩), find a new vector ( vec{w}=3vec{u}+2vec{v}).

Solution

First, we must multiply each vector by the scalar.

[egin{align*} 3u &= 3⟨3,−2⟩ [4pt] &= ⟨9,−6⟩ [4pt] 2v &= 2⟨−1,4⟩ [4pt] &= ⟨−2,8⟩ end{align*}]

Then, add the two together.

[egin{align*} w &= 3u+2v [4pt] &=⟨9,−6⟩+⟨−2,8⟩ [4pt] &= ⟨9−2,−6+8⟩ [4pt] &= ⟨7,2⟩ end{align*}]

So, (w=⟨7,2⟩).

Finding Component Form

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the (x) direction, and the vertical component is the (y) direction. For example, we can see in the graph in Figure (PageIndex{13}) that the position vector (⟨2,3⟩) comes from adding the vectors (v_1) and (v_2). We have (v_2) with initial point ((0,0)) and terminal point ((2,0)).

[egin{align*} v_1 &= ⟨2−0,0−0⟩ [4pt] &= ⟨2,0⟩ end{align*}]

We also have (v_2) with initial point ((0,0)) and terminal point ((0, 3)).

[egin{align*} v_2 &= ⟨0−0,3−0⟩ [4pt] &= ⟨0,3⟩ end{align*}]

Therefore, the position vector is

[egin{align*} v &= ⟨2+0,3+0⟩ [4pt] &= ⟨2,3⟩ end{align*}]

Using the Pythagorean Theorem, the magnitude of (v_1) is (2), and the magnitude of (v_2) is (3). To find the magnitude of (v), use the formula with the position vector.

[egin{align*} |v| &= sqrt{{|v_1|}^2+{|v_2|}^2} [4pt] &= sqrt{2^2+3^2} [4pt] &= sqrt{13} end{align*}]

The magnitude of (v) is (sqrt{13}). To find the direction, we use the tangent function ( an heta=dfrac{y}{x}).

[egin{align*} an heta &= dfrac{v_2}{v_1} [4pt] an heta &= dfrac{3}{2} [4pt] heta &={ an}^{−1}left(dfrac{3}{2} ight)=56.3° end{align*}]

Thus, the magnitude of (vec{v}) is (sqrt{13}) and the direction is (56.3^{circ}) off the horizontal.

Example (PageIndex{6}): Finding the Components of the Vector

Find the components of the vector (vec{v}) with initial point ((3,2)) and terminal point ((7,4)).

Solution

First find the standard position.

[egin{align*} v &= ⟨7−3,4−2⟩ [4pt] &= ⟨4,2⟩ end{align*}]

See the illustration in Figure (PageIndex{14}).

The horizontal component is (vec{v_1}=⟨4,0⟩) and the vertical component is (vec{v_2}=⟨0,2⟩).

Finding the Unit Vector in the Direction of (v)

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude (1). We call a vector with a magnitude of (1) a unit vector. We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as (vec{i}=⟨1,0⟩) and is directed along the positive horizontal axis. The vertical unit vector is written as (vec{j}=⟨0,1⟩) and is directed along the positive vertical axis. See Figure (PageIndex{15}).

THE UNIT VECTORS

If (vec{v}) is a nonzero vector, then (dfrac{v}{| v |}) is a unit vector in the direction of (v). Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.

Example (PageIndex{7}): Finding the Unit Vector in the Direction of (v)

Find a unit vector in the same direction as (v=⟨−5,12⟩).

Solution

First, we will find the magnitude.

[egin{align*} |v| &= sqrt{{(−5)}^2+{(12)}^2} [4pt] &= sqrt{25+144} [4pt] &=sqrt{169} [4pt] &= 13 end{align*}]

Then we divide each component by (| v |), which gives a unit vector in the same direction as (vec{v}):

(dfrac{v}{| v |} = −dfrac{5}{13}i+dfrac{12}{13}j )

or, in component form

(dfrac{v}{| v |}= left langle -dfrac{5}{13},dfrac{12}{13} ight angle)

See Figure (PageIndex{16}).

Verify that the magnitude of the unit vector equals (1). The magnitude of (−dfrac{5}{13}i+dfrac{12}{13}j) is given as

[egin{align*} sqrt{ {left(−dfrac{5}{13} ight)}^2+{ left(dfrac{12}{13} ight) }^2 } &= sqrt{dfrac{25}{169}+dfrac{144}{169}} [4pt] &= sqrt{dfrac{169}{169}} &=1 end{align*}]

The vector (u=dfrac{5}{13}i+dfrac{12}{13}j) is the unit vector in the same direction as (v=⟨−5,12⟩).

Performing Operations with Vectors in Terms of (i) and (j)

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of (i) and (j).

VECTORS IN THE RECTANGULAR PLANE

Given a vector (vec{v}) with initial point (P=(x_1,y_1)) and terminal point (Q=(x_2,y_2)), (vec{v}) is written as

[v=(x_2−x_1)i+(y_1−y_2)j]

The position vector from ((0,0)) to ((a,b)), where ((x_2−x_1)=a) and ((y_2−y_1)=b), is written as (vec{v} = vec{ai}+ vec{bj}). This vector sum is called a linear combination of the vectors (vec{i}) and (vec{j}).

The magnitude of (vec{v} = overrightarrow{ai} + overrightarrow{bj}) is given as (| v |=sqrt{a^2+b^2}). See Figure (PageIndex{17}).

Example (PageIndex{8A}): Writing a Vector in Terms of (i) and (j)

Given a vector (vec{v}) with initial point (P=(2,−6)) and terminal point (Q=(−6,6)), write the vector in terms of (vec{i}) and (vec{j}).

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

[egin{align*} v &= (x_2−x_1)i+(y_2−y_1)j [4pt] &=(−6−2)i+(6−(−6))j [4pt] &= −8i+12j end{align*}]

Example (PageIndex{8B}): Writing a Vector in Terms of (i) and (j) Using Initial and Terminal Points

Given initial point (P_1=(−1,3)) and terminal point (P_2=(2,7)), write the vector (vec{v}) in terms of (vec{i}) and (vec{j}).

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

[egin{align*} v &= (x_2−x_1)i+(y_2−y_1)j [4pt] v &= (2−(−1))i+(7−3)j [4pt] &= 3i+4j end{align*}]

Exercise (PageIndex{3})

Write the vector (vec{u}) with initial point (P=(−1,6)) and terminal point (Q=(7,−5)) in terms of (vec{i}) and (vec{j}).

Answer

(u=8i−11j)

Performing Operations on Vectors in Terms of (i) and (j)

When vectors are written in terms of (i) and (j), we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

ADDING AND SUBTRACTING VECTORS IN RECTANGULAR COORDINATES

Given (v = ai + bj) and (u = ci + dj), then

[egin{align*} v+u &= (a+c)i+(b+d)j [4pt] v−u &= (a−c)i+(b−d)j end{align*}]

Example (PageIndex{9}): Finding the Sum of the Vectors

Find the sum of (v_1=2i−3j) and (v_2=4i+5j).

Solution

[egin{align*} v_1+v_2 &= (2+4)i+(−3+5)j [4pt] &= 6i+2j end{align*}]

Calculating the Component Form of a Vector: Direction

We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using (i) and (j). For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.

Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with (| v |) replacing (r).

VECTOR COMPONENTS IN TERMS OF MAGNITUDE AND DIRECTION

Given a position vector (v=⟨x,y⟩) and a direction angle ( heta),

[ egin{align*} cos heta &= dfrac{x}{|v|} & ext{ and }&& sin heta &=dfrac{y}{|v|} [4pt]
x &= |v| cos heta &&& y &= |v| sin heta end{align*}]

Thus, (v=xi+yj=| v | cos heta i+| v | sin heta j), and magnitude is expressed as (| v |=sqrt{x^2+y^2}).

Example (PageIndex{10}): Writing a Vector in Terms of Magnitude and Direction

Write a vector with length (7) at an angle of (135°) to the positive x-axis in terms of magnitude and direction.

Solution

Using the conversion formulas (x=| v | cos heta i) and (y=| v | sin heta j), we find that

[ egin{align*} x &= 7cos(135°)i [4pt] &= −dfrac{7sqrt{2}}{2} [4pt] y &=7 sin(135°)j [4pt] &= dfrac{7sqrt{2}}{2} end{align*}]

This vector can be written as (v=7cos(135°)i+7sin(135°)j) or simplified as

(v=−dfrac{7sqrt{2}}{2}i+dfrac{7sqrt{2}}{2}j)

Exercise (PageIndex{4})

A vector travels from the origin to the point ((3,5)). Write the vector in terms of magnitude and direction.

Answer

(v=sqrt{34}cos(59°)i+sqrt{34}sin(59°)j)

Magnitude = (34)

( heta={ an}^{−1}left(dfrac{5}{3} ight)=59.04°)

Finding the Dot Product of Two Vectors

As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

DOT PRODUCT

The dot product of two vectors (v=⟨a,b⟩) and (u=⟨c,d⟩) is the sum of the product of the horizontal components and the product of the vertical components.

[v⋅u=ac+bd]

To find the angle between the two vectors, use the formula below.

[cos heta=dfrac{v}{| v |}⋅dfrac{u}{| u |}]

Example (PageIndex{11A}): Finding the Dot Product of Two Vectors

Find the dot product of (v=⟨5,12⟩) and (u=⟨−3,4⟩).

Solution

Using the formula, we have

[egin{align*} v⋅u &= ⟨5,12⟩⋅⟨−3,4⟩ [4pt] &= 5⋅(−3)+12⋅4 [4pt] &= −15+48 [4pt] &= 33 end{align*}]

Example (PageIndex{11B}): Finding the Dot Product of Two Vectors and the Angle between Them

Find the dot product of (v_1 = 5i + 2j) and (v_2 = 3i + 7j). Then, find the angle between the two vectors.

Solution

Finding the dot product, we multiply corresponding components.

[ egin{align*} v_1⋅v_2 &= ⟨5,2⟩⋅⟨3,7⟩ [4pt] &= 5⋅3+2⋅7 [4pt] &= 15+14 [4pt] &= 29 end{align*}]

To find the angle between them, we use the formula (cos heta=dfrac{v}{|v|}⋅dfrac{u}{|u|}).

[egin{align*} dfrac{v}{|v|}cdot dfrac{u}{|u|} &= left langle dfrac{5}{sqrt{29}}+dfrac{2}{sqrt{29}} ight angle cdot left langle dfrac{3}{sqrt{58}}+dfrac{7}{sqrt{58}} ight angle [4pt] &=dfrac{5}{sqrt{29}}cdot dfrac{3}{sqrt{58}}+dfrac{2}{sqrt{29}}cdot dfrac{7}{sqrt{58}} [4pt] &= dfrac{15}{sqrt{1682}}+dfrac{14}{sqrt{1682}} &=dfrac{29}{sqrt{1682}} [4pt] &= 0.707107 [4pt] {cos}^{-1}(0.707107) &= 45° end{align*}]

See Figure (PageIndex{18}).

Example (PageIndex{11C}): Finding the Angle between Two Vectors

Find the angle between (u=⟨−3,4⟩) and (v=⟨5,12⟩).

Solution

Using the formula, we have

[egin{align*} heta &= {cos}^{−1}left(dfrac{u}{|u|}⋅dfrac{v}{|v|} ight) [4pt] left(dfrac{u}{|u|}⋅dfrac{v}{|v|} ight) &= dfrac{−3i+4j}{5}⋅dfrac{5i+12j}{13} [4pt] &= left(− dfrac{3}{5}⋅ dfrac{5}{13} ight)+left(dfrac{4}{5}⋅ dfrac{12}{13} ight) [4pt] &= −dfrac{15}{65}+dfrac{48}{65} [4pt] &= dfrac{33}{65} [4pt] heta &= {cos}^{−1}left(dfrac{33}{65} ight) [4pt] &= 59.5^{circ} end{align*}]

See Figure (PageIndex{19}).

Example (PageIndex{11D}): Finding Ground Speed and Bearing Using Vectors

We now have the tools to solve the problem we introduced in the opening of the section.

An airplane is flying at an airspeed of (200) miles per hour headed on a SE bearing of (140°). A north wind (from north to south) is blowing at (16.2) miles per hour. What are the ground speed and actual bearing of the plane? See Figure (PageIndex{20}).

Solution

The ground speed is represented by (x) in the diagram, and we need to find the angle (alpha) in order to calculate the adjusted bearing, which will be (140°+alpha).

Notice in Figure (PageIndex{20}), that angle (angle BCO) must be equal to angle (angle AOC) by the rule of alternating interior angles, so angle (angle BCO) is 140°. We can find (x) by the Law of Cosines:

[egin{align*} x^2 &= {(16.2)}^2+{(200)}^2−2(16.2)(200) cos(140°) [4pt] x^2 &= 45,226.41 [4pt] x &= sqrt{45,226.41} [4pt] x &= 212.7 end{align*}]

The ground speed is approximately (213) miles per hour. Now we can calculate the bearing using the Law of Sines.

[egin{align*} dfrac{sin alpha}{16.2} &= dfrac{sin(140°)}{212.7} [4pt] sin alpha &= dfrac{16.2 sin(140°)}{212.7} [4pt] &=0.04896 [4pt] {sin}^{−1}(0.04896) &= 2.8° end{align*}]

Therefore, the plane has a SE bearing of (140°+2.8°=142.8°). The ground speed is (212.7) miles per hour.

Media: Access these online resources for additional instruction and practice with vectors.

  • Introduction to Vectors
  • Vector Operations
  • The Unit Vector

Key Concepts

  • The position vector has its initial point at the origin. See Example (PageIndex{1}).
  • If the position vector is the same for two vectors, they are equal. See Example (PageIndex{2}).
  • Vectors are defined by their magnitude and direction. See Example (PageIndex{3}).
  • If two vectors have the same magnitude and direction, they are equal. See Example (PageIndex{4}).
  • Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example (PageIndex{5}).
  • Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See Example (PageIndex{6}) and Example (PageIndex{7}).
  • Vectors are comprised of two components: the horizontal component along the positive (x)-axis, and the vertical component along the positive (y)-axis. See Example (PageIndex{8}).
  • The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.
  • The magnitude of a vector in the rectangular coordinate system is (| v |=sqrt{a^2+b^2}). See Example (PageIndex{9}).
  • In the rectangular coordinate system, unit vectors may be represented in terms of (ii) and (jj) where (i) represents the horizontal component and (j) represents the vertical component. Then, (v = ai + bj) is a scalar multiple of (v) by real numbers (a) and (b). See Example (PageIndex{10}) and Example (PageIndex{11}).
  • Adding and subtracting vectors in terms of (i) and (j) consists of adding or subtracting corresponding coefficients of (i) and corresponding coefficients of (j). See Example (PageIndex{12}).
  • A vector (v = ai + bj) is written in terms of magnitude and direction as (v=| v |cos heta i+| v |sin heta j). See Example (PageIndex{13}).
  • The dot product of two vectors is the product of the (i) terms plus the product of the (j) terms. See Example (PageIndex{14}).
  • We can use the dot product to find the angle between two vectors. Example (PageIndex{15}) and Example (PageIndex{16}).
  • Dot products are useful for many types of physics applications. See Example (PageIndex{17}).

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The value of (overrightarrow>+overrightarrow>+overrightarrow>+overrightarrow>) is ………………

Solution:
(c) (overrightarrow<0>)

Question 3.
The unit vector parallel to the resultant of the vectors (hat+hat-hat) and (hat-2 hat+hat) is ………………

Solution:

Question 4.
A vector (overrightarrow) makes 60° and 45° with the positive direction of the x and y axes respectively. Then the angle between (overrightarrow) and the z-axis is …………….
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Solution:
(b) 60°
α = 60°, β = 45°
We know cos 2 α + cos 2 β + cos 2 γ = 1
(i.e.,) (left(frac<1><2> ight)^<2>+left(frac<1>> ight)^<2>) + cos 2 γ = 1
cos 2 γ = 1 – (frac<1><4>-frac<1><2>=frac<1><4>)
cos γ = (frac<1><2>) ⇒ y = π/3 = 60°

Question 5.
If (overrightarrow=3 hat+2 hat+hat) and the position vector of B is (hat+3 hat-hat), then the position vector A is …………………

Solution:

Question 6.
A vector makes equal angle with the positive direction of the coordinate axes. Then each angle is equal to …………..

Solution:

Question 8.
If ABCD is a parallelogram, then (overrightarrow+overrightarrow+overrightarrow+overrightarrow) is equal to ……………

Solution:

Question 9.
One of the diagonals of parallelogram ABCD with (vec) and (vec) as adjacent sides is (vec+vec). The other diagonal (overrightarrow>) is ……………

Solution:

Question 12.
If (vec=frac<9 vec+7 vec><16>), then the point P whose position vector (vec) divides the line joining the points with position vectors (vec) and (vec) in the ratio ………………
(a) 7 : 9 internally
(b) 9 : 7 internally
(c) 9 : 7 externally
(d) 7 : 9 externally
Solution:

Question 13.
If (lambda hat+2 lambda hat+2 lambda hat) is a unit vector, then the value of λ is ……………..
(a) (frac<1><3>)
(b) (frac<1><4>)
(c) (frac<1><9>)
(d) (frac<1><2>)
Solution:

Question 14.
Two vertices of a triangle have position vectors (3 hat+4 hat-4 hat) and (2 hat+3 hat+4 hat) .If the position vector of the centroid is (hat+2 hat+3 hat), then the position vector of the third vertex is ………………….

Solution:

Question 15.

(a) 42
(b) 12
(c) 22
(d) 32
Solution:
(c) 22

Question 17.
The value of θ ∈ (0, (frac<2>)) for which the vectors are perpendicular, is equal to …………………

Solution:

Question 18.

(a) 15
(b) 35
(c) 45
(d) 25
Solution:
(d) 25

Question 21.
If the projection of (5hat -hat-3 hat) on the vector (hat+3 hat+lambda hat) is same as the projection of (hat+3 hat+lambda hat) on (5hat- hat-3 hat) then λ is equal to ………………
(a) ±4
(b) ±3
(c) ±5
(d) ±1
Solution:
(c) ±5

Question 22.
If (1, 2, 4) and (2, – 3λ, – 3) are the initial and terminal points of the vector (hat+5 hat-7 hat), then the value of λ is equal to ……………..
(a) (frac<7><3>)
(b) (-frac<7><3>)
(c) (-frac<5><3>)
(d) (frac<5><3>)
Solution:

Question 23.
If the points whose position vector (10 hat+3 hat, 12 hat-5 hat) and (vec hat+11 hat) are collinear then a is equal to ………………
(a) 6
(b) 2
(c) 5
(d) 8
Solution:
(d) 8

equating (hat) components
⇒ -8 = 8t ⇒ t = -1
equation (hat) components
t(a – 10) = 2
(i.e.,) (-1) (a – 10) = 2
a – 10 = -2
a = – 2 + 10 = -8

Question 24.
If then x is equal to …………..
(a) 5
(b) 7
(c) 26
(d) 10
Solution:
(c) 26


10.8: Vectors - Mathematics

Prelude: A vector, as defined below, is a specific mathematical structure. It has numerous physical and geometric applications, which result mainly from its ability to represent magnitude and direction simultaneously. Wind, for example, had both a speed and a direction and, hence, is conveniently expressed as a vector. The same can be said of moving objects and forces. The location of a points on a cartesian coordinate plane is usually expressed as an ordered pair (x, y), which is a specific example of a vector. Being a vector, (x, y) has a a certain distance (magnitude) from and angle (direction) relative to the origin (0, 0). Vectors are quite useful in simplifying problems from three-dimensional geometry.

Definition: A scalar, generally speaking, is another name for "real number."

Definition: A vector of dimension n is an ordered collection of n elements, which are called components.

Notation: We often represent a vector by some letter, just as we use a letter to denote a scalar (real number) in algebra. In typewritten work, a vector is usually given a bold letter, such as A, to distinguish it from a scalar quantity, such as A. In handwritten work, writing bold letters is difficult, so we typically just place a right-handed arrow over the letter to denote a vector. An n-dimensional vector A has n elements denoted as A1, A2, . An. Symbolically, this can be written in multiple ways:

A = <A1, A2, . An>
A = (A1, A2, . An)

Example: (2,-5), (-1, 0, 2), (4.5), and (PI, a, b, 2/3) are all examples of vectors of dimension 2, 3, 1, and 4 respectively. The first vector has components 2 and -5.

Note: Alternately, an "unordered" collection of n elements is called a "set."

Definition: Two vectors are equal if their corresponding components are equal.

Example: If A = (-2, 1) and B = (-2, 1), then A = B since -2 = -2 and 1 = 1. However, (5, 3) not_equal (3, 5) because even though they have the same components, 3 and 5, the component do not occur in the same order. Contrast this with sets, where <5, 3>= <3, 5>.

Definition: The magnitude of a vector A of dimension n, denoted |A|, is defined as

Geometrically speaking, magnitude is synonymous with "length," "distance", or "speed." In the two-dimensional case, the point represented by the vector A = (A1, A2) has a distance from the origin (0, 0) of sqrt(A1^2 + A2^2) according to the pythagorean theorem. In the three-dimension case, the point represented by the vector A = (A1, A2, A3) has a distance from the origin of sqrt(A1^2 + A2^2 + A3^2) according to the three-dimensional form of the Pythagorean theorem (A box with sides a, b, and c has a diagonal of length sqrt(a2+b2+c2) ). With vectors of dimension n greater than three, our geometric intuition fails, but the algebraic definition remains.

Definition: The sum of two vectors A = (A1, A2, . An) and B = (B1, B2, . Bn) is defined as

A + B = (A1 + B1, A2 + B2, . An + Bn)

Note: Addition of vectors is only defined if both vectors have the same dimension.

Justification: Physical and geometric applications warrant such a definition. IF a train travels East at 5 meters/second relative to the ground, which will be denoted in vector notation as VT = (0, 5), and a person on the train walks South at 1 meter/second relative to the train, which will be denoted as VP = (-1, 0), THEN the direction and speed that the person is traveling relative to the ground is represented by the vector VG = VT + VP = (0, 5) + (-1, 0) = (0 + -1, 5 + 0) = (-1, 5). This vector has a magnitude of |VG| = sqrt((-1)^2 + 5^2) = sqrt(26) = 5.099. which means that the person is traveling at about 5.099 meters/second relative to the ground and the net direction is mostly East but slightly South.

Definition: The scalar product of a scalar k by a vector A = (A1, A2, . An) is defined as

Note: In general, 0A = (0, 0, . 0) and 1A = A, just as in the algebra of scalars. The vector of any dimension n with all zero elements (0, 0, . 0) is called the zero vector and is denoted 0.


Regents Physics - Math Review

Review required concepts and skills required for success in the NY Regents Physics course.

  1. Express answers with correct values with respect to significant figures.
  2. Use scientific notation to express physical values efficiently.
  3. Convert and estimate SI units.
  4. Differentiate between scalar and vector quantities.
  5. Use scaled diagrams to represent and manipulate vector quantities.
  6. Determine x- and y-components of two-dimensional vectors.
  7. Determine the angle of a vector given its components.

Significant Figures

Significant Figures (or sig figs, for short) represent a manner of showing which digits in a number are known to some level of certainty. But how do we know which digits are significant? There are some rules to help us with this. If we start with a number in scientific notation:

  • All non-zero digits are significant.
  • All digits between non-zero digits are significant.
  • Zeroes to the left of significant digits are not significant.
  • Zeroes to the right of significant digits are significant.

When you make a measurement in physics, you want to write what you measured using significant figures. To do this, write down as many digits as you are absolutely certain of, then take a shot at one more digit as accurately as you can. These are your significant figures. On the Regents Physics Exam, you may answer any problem by showing three or four significant figures.

Scientific Notation

Although physics and mathematics aren&rsquot the same thing, they are in many ways closely related. Just like English is the language of this content, mathematics is the language of physics. A solid understanding of a few simple math concepts will allow us to communicate and describe the physical world both efficiently and accurately.

Because measurements of the physical world vary so tremendously in size (imagine trying to describe the distance across the United States in units of hair thicknesses), physicists often times use what is known as scientific notation to denote very large and very small numbers. These very large and very small numbers would become quite cumbersome to write out repeatedly. Imagine writing 4,000,000,000,000 over and over again. Your hand would get tired and your pen would rapidly run out of ink! Instead, it&rsquos much easier to write this number as 4×10 12 . See how much easier that is? Or on the smaller scale, the thickness of the insulating layer (known as a gate dielectric) in the integrated circuits that power our computers and other electronics can be less than 0.000000001 m. It&rsquos easy to lose track of how many zeros you have to deal with, so scientists instead would write this number as 1×10 -9 m. See how much simpler life can be with scientific notation?

To properly use scientific notation, just follow these simple rules. Start by showing all the significant figures in the number you&rsquore describing, with the decimal point after the first significant digit. Then, show your number being multiplied by 10 to the appropriate power in order to give you the correct value.

It sounds more complicated than it is. Let&rsquos say, for instance, you want to show the number 300,000,000 in scientific notation (a very useful number in physics), and let&rsquos assume we know this value to three significant digits. We would start by writing our three significant digits, with the decimal point after the first digit, as &ldquo3.00&rdquo. Now, we need to multiple this number by 10 to some power in order to get back to our original value. In this case, we multiple 3.00 by 10 8 , for an answer of 3.00×10 8 . Interestingly, the power you raise the 10 to is exactly equal to the number of digits you moved the decimal to the left as you converted from standard to scientific notation. Similarly, if you start in scientific notation, to convert to standard notation, all you have to do is remove the 10 8 power by moving the decimal point eight digits to the right. Presto, you&rsquore an expert in scientific notation!

But, what do you do if the number is much smaller than zero? Same basic idea… let&rsquos assume we&rsquore dealing with the approximate radius of an electron, which is 0.00000000000000282 m. It&rsquos easy to see how unwieldy this could become. We can write this in scientific notation by writing our three significant digits, with the decimal point after the first digit, as &ldquo2.82.&rdquo Again, we multiple this number by some power to 10 in order to get back to our original value. Because our value is less than 1, we need to use negative powers of 10. If we raise 10 to the power -15, specifically, we get a final value of 2.82×10 -15 m. In essence, for every digit we moved the decimal place, we add another power of 10. And if we start with scientific notation, all we do is move the decimal place left one digit for every negative power of 10.

Metric System

Physics involves the study, prediction, and analysis of real-world phenomena. To communicate data accurately, we must set specific standards for our basic measurements. The physics community has standarized on what is known as the Système International (SI), which defines seven baseline measurements and their standard units, forming the foundation of what is called the metric system of measurement. The SI system is oftentimes referred to as the mks system, as the three most common measurement units are meters, kilograms, and seconds, which we'll focus on for majority of this course. The fourth SI base unit we'll use in this course, the ampere, will be introduced in the current electricity section.

The base unit of length in the metric system, the meter, is roughly equivalent to the U.S. yard. For smaller measurements, the meter is divided up into 100 parts, known as centimeters, and each centimeter is made up of 10 millimeters. For larger measurements, the meter is grouped into larger units of 1000 meters, known as a kilometer. The length of a baseball bat is approximately one meter, the radius of a U.S. quarter is approximately a centimeter, and the diameter of the metal in a wire paperclip is roughly one millimeter.

The base unit of mass, the kilogram, is roughly equivalent to two U.S. pounds. A cube of water 10cm × 10cm × 10cm has a mass of 1 kilogram. Kilograms can also be broken up into larger and smaller units, with commonly used measurements of grams (1/1000th of a kilogram) and milligrams (1/1000th of a gram). The mass of a textbook is approximately 2 to 3 kilograms, the mass of a baseball is approximately 145 grams, and the mass of a mosquito is 1 to 2 milligrams.

The base unit of time, the second, is likely already familiar. Time can also be broken up into smaller units such as milliseconds (10 -3 seconds), microseconds (10 -6 seconds), and nanoseconds (10 -9 seconds), or grouped into larger units such as minutes (60 seconds), hours (60 minutes), days (24 hours), and years (365.25 days).

The metric system is based on powers of 10, allowing for easy conversion from one unit to another. The front page of the Physics Reference Table includes a chart showing the meaning of the commonly used metric prefixes, which can be extremely valuable in performing unit conversions.

Converting from one unit to another can be accomplished in a straightforward manner if you follow the described procedure:

  1. Draw a horizontal and vertical line to create four quadrants.
  2. Place your initial measurement with units in the upper left-hand quadrant.
  3. Copy the units from the upper left-hand quadrant to the lower right-hand quadrant.
  4. Write the units you want to convert to in the upper right-hand quadrant.
  5. For any units on the top right-hand quadrant with a prefix, use the reference table to determine the notation for that prefix. Write that prefix in the bottom right-hand quadrant.
  6. For any units on the bottom right-hand quadrant with a prefix, write the notation for that prefix in the top right-hand quadrant.
  7. Multiple through the problem, taking care to accurately record units. You should be left with your final answer in the desired units.

Let's take a look at a couple examples:

Algebra and Trigonometry

Just as we find the English language a convenient tool for conveying thoughts to each other, we need a convenient language for conveying our understanding of the world around us in order to understand its behavior. The language most commonly (and conveniently) used to describe the natural world is mathematics. Therefore, to understand physics, we need to be fluent in the mathematics of the topics we&rsquoll study in this course… specifically basic algebra and trigonometry.

Now don&rsquot you fret or frown, for those whom the word &ldquotrig&rdquo conjures up feelings of pain, angst, and frustration, have no fear. We will need only the most basic of algebra and trigonometry knowledge in order to successfully solve a wide range of physics problems.

A vast majority of problems requiring algebra can be solved using the same problem solving strategy. First, analyze the problem and write down what you know, what you need to find, and make a picture or diagram to better understand the problem if it makes sense. Then, start your solution by selecting an appropriate formula from the Reference Table (a comprehensive formula sheet covering material for the entire course). Next, substitute given information from the problem statement into the formula (with units). Finally, solve the problem for your final answer, making sure to show your answer with units. This strategy, known as the AFSA strategy, can be summarized as:

  • Analysis – Write down what is given, what you're asked to find, and make a diagram, if appropriate.
  • Formula – Copy the appropriate formula from the reference table.
  • Substitution (with units) – Replace variables in the selected formula with data from the problem statement.
  • Answer (with units) – Solve the problem and record your final answer with units. Place a box around your final answer.

Our use of trigonometry, the study of triangles, can be distilled down into the definitions of the three basic trigonometric functions. If you can use the definitions of the sine, cosine, and tangent, you&rsquoll be fine in this course. Even better – if you&rsquore a little rusty on the basic trig definitions, they&rsquore all included for you on the Regents Physics Reference Table – you don&rsquot even have to memorize them.

Remember to use the AFSA strategy and your reference table and you&rsquoll be well prepared for anything the Regents Physics Exam may throw at you.

Vectors and Scalars

Quantities in physics are used to represent real-world measurements, and therefore physicists use these quantities as tools to better understand the world. In examining these quantities, there are times when just a number, with a unit, can completely describe a situation. These numbers, which have a magnitude, or size, only are known as scalars. Examples of scalars include quantities such as temperature, mass, and time. At other times, a quantity is more descriptive if it also includes a direction. These quantities which have both a magnitude and direction are known as vectors. Vector quantities you may be familiar with include force, velocity, and acceleration.

Most students will be familiar with scalars, but to many, vectors may be a new and confusing concept. By learning just a few rules for dealing with vectors, though, you&rsquoll find that they are a powerful tool for problem solving.

Vectors are often represented as arrows, with the length of the arrow indicating the magnitude of the quantity, and the direction of the arrow indicating the direction of the vector. In the figure at right, vector B has a magnitude greater than that of vector A. Vectors A and B point in the same direction, however. It&rsquos also important to note that vectors can be moved anywhere in space. The positions of A and B could be reversed, and the individual vectors would retain their values of magnitude and direction. This makes adding vectors very straightforward!

To add vectors A and B, all we have to do is line them up so that the tip of the first vector touches the tail of the second vector. Then, to find the sum of the vectors, known as the resultant, all we have to do is draw a straight line from the start of the first vector to the end of the last vector. This method works with any number of vectors.

So then, how do we subtract two vectors? Let&rsquos try it by subtracting B from A. We could rewrite the expression A - B as A + -B. Now it becomes an addition problem, we just have to figure out how to express –B. This is easier than it sounds – to find the opposite of a vector, we just point the vector in the opposite direction. Therefore, we can use what we already know about the addition of vectors to find the resultant of A-B.

Components of Vectors

We&rsquoll learn more about vectors as we go, but before we move on, there&rsquos one more skill we need to learn. Vectors at angles can be challenging to deal with. By transforming a vector at an angle into two vectors, one parallel to the x-axis and one parallel to the y-axis, we can greatly simplify problem solving. To break a vector up into its components, we can use our basic trig functions. To help us out even further, the Regents Physics Reference Table includes the exact formulas we need to determine the x- and y-components of any vector if we know that vector&rsquos magnitude and direction.

In similar fashion, we can use the components of a vector in order to build the original vector. Graphically, if we line up the component vectors tip-to-tail, the original vector runs from the starting point of the first vector to the ending point of the last vector. To determine the magnitude of the resulting vector algebraically, just apply the Pythagorean Theorem!


10.8: Vectors - Mathematics

This video and text below takes a look at Vectors and Scalars.

The magnitude of vector x is written as |x|.

The magnitude of vector is written as |AB|.

A vector with magnitude 0 is called the zero vector , written 0. A vector with magnitude 1 is called a unit vector .

Vectors are equal if they have the same magnitude and the same direction.

The inverse of a vector is a vector of equal magnitude but in the opposite direction. The inverse of is - or and the inverse of a is -a.

Scalars have magnitude but not direction. Vectors can be multiplied by a scalar to produce another vector.

Multiplying vector x by 3 will give a new vector 3 times the length and parallel to x.

When 2 vectors are added or subtracted the vector produced is called the resultant.

The resultant is identified by a double arrowhead.

To add two vectors you apply the first vector and then the second.

+ =

Subtracting a vector is the same as adding its inverse.

Moving from A to C through B is the same as moving through D.

+ = + =


Vector Math in R

Vector Math means we should be able to perform mathematical operations on vectors in R.

Let’s define two vectors and perform the addition operation.

Output

The output here is the vector, which is the sum of the corresponding elements of two vectors vcA and vcB.

Subtraction in R Vector

We can perform the subtraction the same way we have performed the vector addition.

Output

Division in R Vector

To perform division operation in R vector, use the following code.

Output


10.8: Vectors - Mathematics

There are many situations when we might wish to know whether a set of vectors is linearly dependent, that is if one of the vectors is some combination of the others.

Two vectors u and v are linearly independent if the only numbers x and y satisfying x u +y v =0 are x=y=0. If we let

then x u +y v =0 is equivalent to

If u and v are linearly independent, then the only solution to this system of equations is the trivial solution, x=y=0. For homogeneous systems this happens precisely when the determinant is non-zero. We have now found a test for determining whether a given set of vectors is linearly independent: A set of n vectors of length n is linearly independent if the matrix with these vectors as columns has a non-zero determinant. The set is of course dependent if the determinant is zero.

Example

The vectors <1,2> and <-5,3> are linearly independent since the matrix

has a non-zero determinant.

Example

The vectors u =<2,-1,1>, v =<3,-4,-2>, and w =<5,-10,-8> are dependent since the determinant

is zero. To find the relation between u , v , and w we look for constants x, y, and z such that

This is a homogeneous system of equations. Using Gaussian Elimination, we see that the matrix

Thus, y=-3z and 2x=-3y-5z=-3(-3z)-5z=4z which implies 0=x u +y v +z w =2z u -3z v +z w or equivalently w =-2 u +3 v . A quick arithmetic check verifies that the vector w is indeed equal to -2 u +3 v .


The numerator of the ratio 10.8:10.8 contains 1 decimal and the denominator contains 1 decimal

The lowest possible whole number equivalent ratio of the ratio 10.8:10.8 is:

If you wish to express the ratio 10.8:10.8 as n to 1 then the ratio would be:

If you wish to express the ratio 10.8:10.8 as 1 to n then the ratio would be:

The ratio 10.8:10.8 expressed as a fraction is [calculated using the ratio to fraction calculator]:

The ratio 10.8:10.8 expressed as a percentage is [calculated using the ratio to percentage calculator]:


Vectors

Vector Notation: The lower case letters a-h, l-z denote scalars. Uppercase bold AZ denote vectors. Lowercase bold i, j, k denote unit vectors. <a, b>denotes a vector with components a and b. <x1, . xn>denotes vector with n components of which are x1, x2, x3, . xn. |R| denotes the magnitude of the vector R.

|<a, b>| = magnitude of vector = (a 2 + b 2 )

<a, b> + <c, d> = <a+c, b+d>

k <a, b> = <ka, kb>

<a, b> <c, d> = ac + bd

R S= |R| |S| cos ( = angle between them)

R S= S R

(a R) (bS) = (ab) R S

R (S + T)= R S+ R T

|R x S| = |R| |S| sin ( = angle between both vectors). Direction of R x S is perpendicular to A & B and according to the right hand rule.

S x R = – R x S

(a R) x S = R x (a S) = a (Rx S)

R x (S + T) = R x S + Rx T

If a, b, c = angles between the unit vectors i, j,k and R Then the direction cosines are set by:

COs a = (R i) / |R| COs b = (R j) / |R| COs c = (R k) / |R|

|R x S| = Area of parrallagram with sides Rand S.

Component of R in the direction of S = |R|COs = (R S) / |S|(scalar result)

Projection of R in the direction of S = |R|COs = (R S) S/ |S| 2 (vector result)

Prelude: A vector, as defined below, is a specific mathematical structure. It has numerous physical and geometric applications, which result mainly from its ability to represent magnitude and direction simultaneously. Wind, for example, had both a speed and a direction and, hence, is conveniently expressed as a vector. The same can be said of moving objects and forces. The location of a points on a cartesian coordinate plane is usually expressed as an ordered pair (x, y), which is a specific example of a vector. Being a vector, (x, y) has a a certain distance (magnitude) from and angle (direction) relative to the origin (0, 0). Vectors are quite useful in simplifying problems from three-dimensional geometry.

Definition: A scalar, generally speaking, is another name for “real number.”

Definition: A vector of dimension n is an ordered collection of n elements, which are called components.

Notation: We often represent a vector by some letter, just as we use a letter to denote a scalar (real number) in algebra. In typewritten work, a vector is usually given a bold letter, such as A, to distinguish it from a scalar quantity, such as A. In handwritten work, writing bold letters is difficult, so we typically just place a right-handed arrow over the letter to denote a vector. An n-dimensional vector A has n elements denoted as A1, A2, …, An. Symbolically, this can be written in multiple ways:

Example: (2,-5), (-1, 0, 2), (4.5), and (PI, a, b, 2/3) are all examples of vectors of dimension 2, 3, 1, and 4 respectively. The first vector has components 2 and -5.

Note: Alternately, an “unordered” collection of n elements is called a “set.”

Definition: Two vectors are equal if their corresponding components are equal.

Example: If A = (-2, 1) and B = (-2, 1), then A = B since -2 = -2 and 1 = 1. However, (5, 3) not_equal (3, 5) because even though they have the same components, 3 and 5, the component do not occur in the same order. Contrast this with sets, where <5, 3>= <3, 5>.

Definition: The magnitude of a vector A of dimension n, denoted |A|, is defined as

Geometrically speaking, magnitude is synonymous with “length,” “distance”, or “speed.” In the two-dimensional case, the point represented by the vector A = (A1, A2) has a distance from the origin (0, 0) of sqrt(A1^2 + A2^2) according to the pythagorean theorem. In the three-dimension case, the point represented by the vector A = (A1, A2, A3) has a distance from the origin of sqrt(A1^2 + A2^2 + A3^2) according to the three-dimensional form of the Pythagorean theorem (A box with sides a, b, and c has a diagonal of length sqrt(a2+b2+c2) ). With vectors of dimension n greater than three, our geometric intuition fails, but the algebraic definition remains.

Definition: The sum of two vectors A = (A1, A2, …, An) and B = (B1, B2, …, Bn) is defined as

A + B = (A1 + B1, A2 + B2, …, An + Bn)

Note: Addition of vectors is only defined if both vectors have the same dimension.

Justification: Physical and geometric applications warrant such a definition. IF a train travels East at 5 meters/second relative to the ground, which will be denoted in vector notation as VT = (0, 5), and a person on the train walks South at 1 meter/second relative to the train, which will be denoted as VP = (-1, 0), THEN the direction and speed that the person is traveling relative to the ground is represented by the vector VG = VT + VP = (0, 5) + (-1, 0) = (0 + -1, 5 + 0) = (-1, 5). This vector has a magnitude of |VG| = sqrt((-1)^2 + 5^2) = sqrt(26) = 5.099…, which means that the person is traveling at about 5.099 meters/second relative to the ground and the net direction is mostly East but slightly South.

Definition: The scalar product of a scalar k by a vector A = (A1, A2, …, An) is defined as

Note: In general, 0A = (0, 0, …, 0) and 1A = A, just as in the algebra of scalars. The vector of any dimension n with all zero elements (0, 0, …, 0) is called the zero vector and is denoted 0


Examples

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Watch the video: MAT 122 Trigonomety Section Part 2 Position Vector (October 2021).