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1.1 Vectors
A vector is a quantity that has both magnitude and direction.
Vectors are drawn as directed line segments and are denoted by boldface letters a or by ~a.
The magnitude of a vector a is its length, denoted |a|.
In practice, a two-dimensional vector is an ordered pair a =< a1 , a2 > of real numbers. The numbers a1
and a2 are called the components of a.
Two vectors are equal if they have the same magnitude and direction. So, it doesn’t matter where the vector
is as long as it has the same magnitude and direction (the same displacement).
A vector < a1 , a2 > with initial point at the origin is called the position vector of the point (a1 , a2 ). A
vector with initial point at the origin is said to be in standard position.
Magnitude: The magnitude (or length) of a vector a=< a1 , a2 > is |a| =
p
(a1 )2 + (a2 )2
Given the points A(x1 , y1 ) and B(x2 , y2 ), the vector a with initial point A and terminal point B (also written
−−
→
AB) is
a =< x2 − x1 , y2 − y1 >
−
−→
The length of the vector AB from A(x1 , y1 ) to B(x2 , y2 ) is
−−
→
Example: Given the points A(−2, 5) and B(4, 1), find the vector AB and its magnitude.
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Math 171, Benjamin
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The zero vector, 0, is the vector < 0, 0 >. It has magnitude 0 and no direction.
Vector Addition: If a =< a1 , a2 > and b =< b1 , b2 >, then the vector a + b is < a1 + b1 , a2 + b2 >
Scalar Multiplication: Multiplying a vector by a scalar (number) changes the magnitude of the vector by
this factor. A negative scalar changes the magnitude and also reverses the direction.
If c is a scalar and a =< a1 , a2 >, then the vector ca =< ca1 , ca2 >.
Two vectors a and b are parallel if b = ca for some scalar c.
Vector Difference: If a =< a1 , a2 > and b =< b1 , b2 >, then the vector a − b is < a1 − b1 , a2 − b2 >
Example: If a =< 3, −4 > and b =< −1, 2 >, find |2a − b|.
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Show that the vectors a =< x − 2, 4y + 3 > and b =< 6 − 3x, −12y − 9 > are parallel.
Example: Consider the triangle ABC shown below where P and Q are the midpoints of AB and BC
−→
−−→
respectively. Show that P Q = 21 AC.
B
P
A
Q
C
Unit Vector: A vector with length (magnitude) 1 is called a unit vector.
A unit vector in the direction of a is:
Example: If a =< −2, 3 >, find a unit vector in the direction of a.
Example: Find a vector of length 5 in the same direction as a.
Basis Vectors: There are two special unit vectors that we use all the time:
i =< 1, 0 >
j =< 0, 1 >
i is the unit vector in the positive horizontal (x) direction, and j is the unit vector in the positive vertical
(y) direction.
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Math 171, Benjamin
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The vectors i and j are called basis vectors because EVERY vector a=< a1 , a2 > can be written in terms
of these unit vectors by
a = a1 i + a2 j
Show this:
Direction: The direction of a vector is the positive angle θ formed by the positive x-axis and the vector.
Suppose you are given a vector a with magnitude |a| and direction θ. Then
a =< |a| cos θ, |a| sin θ >
Example: A vector a has |a|=4 and θ = 60◦ . Write a in component form.
Example: Find the vector a that has |a| = 10 and makes an angle of 330◦ with the positive x-axis.
Example: Find the direction of the vector a = −4i − 5j.
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Applications
The velocity of an object can be modeled by a vector, where the direction of the vector is the direction of
motion, and the magnitude of the vector is the speed.
Example: A jet is flying through a wind that is blowing with a speed of 40 mph in the direction 120◦ from
the positive x-axis. The jet has an airspeed (speed in still air) of 500 mph, and the pilot heads the jet in a
direction 45◦ from the positive x-axis.. Find the ground speed and true course of the jet.
Example: A person wants to swim to a point on the other side of a river directly north from where they
start. If the water is flowing due west at 2 ft/s and the person can swim at 4 ft/s, in what direction does
the person need to head to reach their destination?
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Math 171, Benjamin
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Another application of vectors involves forces. A force has magnitude (measured in pounds or Newtons)
and a direction. If several forces are acting on an object, the resultant force is the vector sum of all these
forces.
Example: Two forces F1 and F2 are acting on an object at a point P . F1 has a magnitude of 40 lbs in
a direction of 60◦ from the positive x-axis. F2 has a magnitude of 20 lbs in a direction of 330◦ from the
positive x-axis. Find the resultant force F along with both its magnitude and direction.
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Math 171, Benjamin
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1.2 The Dot Product
The dot product of two nonzero vectors a and b, denoted a · b, is the number
a · b = |a||b| cos θ
where θ is the angle between a and b, 0 ≤ θ ≤ π. If either vector is the vector 0, then the dot product is 0.
If the vectors are given in component form where a =< a1 , a2 > and b =< b1 , b2 >, then
a · b = a1 b1 + a2 b2
This formula is proved in Section 1.2 of your textbook using the Law of Cosines.
Important: The dot product of two vectors is always a SCALAR, not a vector. For this reason, the dot
product is sometimes called the scalar product.
Example: Calculate a · b in the following scenarios.
(a) a =< 4, 7 >, b =< 3, −3 >
(b) |a| = 7, |b| = 2, and the angle between the two vectors is
π
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3 cases:
Two vectors are orthogonal or perpendicular if if the angle between them is
π
2
or 90◦ .
Thus, two vectors are orthogonal if and only if a · b = 0, since cos π2 = 0.
Example: Find the value(s) of x such that the vectors a =< 2x2 , −1 > and b = 2i − 6xj are orthogonal.
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Orthogonal Complement: Sometimes it is useful to find a vector that is orthogonal to a given vector with
the same length. Given the nonzero vector a =< a1 , a2 >, the orthogonal complement of a is the vector
a⊥ =< −a2 , a1 >
There is another vector that is orthogonal to a: < a2 , −a1 >, but this one doesn’t have a special name.
Find the orthogonal complement of the vector a = −10i + 6j.
Find a vector of length 2 that is orthogonal to a.
Properties of the Dot Product: If a, b, and c are vectors and c is a scalar, then
1. a · a = |a|2
2. a · b = b · a
3. a · (b + c) = a · b + a · c
4. (ca) · b = c(a · b) = a · (cb)
5. 0 · a = 0
Proof of (1):
Proof of (3):
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The dot product can also be used to find the angle between two vectors when they are given in component
form. Since a · b = |a||b| cos θ, then
a·b
cos θ =
|a||b|
a·b
θ = arccos
|a||b|
Example: Given a triangle with vertices A(1, 5), B(2, 1), and C(4, 6), find 6 BCA.
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Math 171, Benjamin
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Work: Work is an application of the dot product to physics. The work done in moving an object through
a distance d by a force F is W = F d. However, this formula only works if the force is applied in the same
direction as the motion. We now generalize to the case where the force may not be applied in the same
direction by using vectors.
−−→
Suppose a force F moves an object from a point P to a point Q. The displacement vector is D = P Q.
How much work is done?
W = F · D = |F||D| cos θ
If the force is
then the unit
If the force is
then the unit
measured in Newtons and the displacement is measured in meters,
of work is N·m, which is also call a joule (J).
measured in pounds and the displacement is measured in feet,
of work is the ft· lb.
Example: A force F = −4i + 2j moves an object from the point P (3, 6) to the point Q(2, 9). How much
work is done if force is measured in lbs and distance is measured in ft?
Example: A horizontal force of 30 N is used to push a box up a 20 m ramp which is inclined at an angle of
15◦ . How much work is done?
Example: If the box is pulled up this same ramp at an angle of 40◦ to the ramp, how much work is done?
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Math 171, Benjamin
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Projections:
Scalar Projection of b onto a (also called the Component of b along a): compa b =
Vector Projection of b onto a: proja b =
a·b
|a|
a·b
|a|
a·b
a
=
a
|a|
|a|2
Example: Find the scalar and vector projections of the vector < 3, −1 > onto the vector < −2, 5 >.
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Math 171, Benjamin
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Example: Find the distance from the point (3, 6) to the line y = 12 x + 1.
Example: Show that |a − b|2 = |a|2 + |b|2 − 2|a||b| cos θ where θ is the angle between the vectors.
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Math 171, Benjamin
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1.3 Vector Functions
Parametric Curves: Sometimes, instead of representing a curve using just x and y, it is more convenient
to use parametric equations using a parameter such as t. This means that the values of x and y are
defined as functions of this parameter, x(t) and y(t).
Example: Sketch the curve represented by the parametric equations x = −3t + 1, y = 2t − 1.
If given a set of parametric equations, it may be useful to convert back into a Cartesian equation (using x
and y only). In order to do this, you must eliminate the parameter. How?
1. If possible, solve one of the parametric equations for t and use substitution.
2. If the parametric equations involve trig functions, use a trig identity, often sin2 θ + cos2 θ = 1.
Example: Eliminate the parameter from the previous example and write a Cartesian equation for the curve.
Sometimes, there may be a restriction on the values of t or the values of x and y may have bounds you need
to watch out for.
If in the previous example, x = −3t + 1, y = 2t − 1, we also had the restriction −1 ≤ t < 2, what does the
curve look like?
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Math 171, Benjamin
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Example: Eliminate the parameter to find a Cartesian equation for the following curves, sketch a graph,
and describe the direction of motion as t increases.
√
(a) x = t, y = 2 − t
(b) x = cos θ, y = sin θ
(c) x = 3 + sin θ, y = 2 + cos θ
Vector Functions: We can define vector functions using these parametric equations by r(t) =< x(t), y(t) >.
It is called a vector function because it takes values of t and produces vectors. These vectors are tracing
out the curve.
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Example: Sketch the curve represented by the vector function r(t) = (4 cos t)i + (sin t)j, 0 ≤ t ≤ π.
Example: The position of an object at t seconds is given by the vector function r(t) =< t2 , t − 2 >.
1. What is the position of the object at time t = 6?
2. At what time, if ever, is the object at position (9, 1)?
3. Does the object pass through the point (49, 7)?
4. Find an equation in x and y whose graph is the path of the object.
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Vector Equation of a Line: If P0 (x0 , y0 ) is a point on the line with position vector r0 =< x0 , y0 > and
v =< a, b > is a vector parallel to a line, then the vector equation of the line is r(t) = r0 + tv.
Parametric equations of the line that passes through the point P (x0 , y0 ) and is parallel to the vector < a, b >
are given by
x = x0 + at
y = y0 + bt
Example: Find a vector equation and parametric equations for the line that passes through the points (2, 5)
and (−1, 7).
Example: Find parametric equations for the line that passes through the point (3, −2) and is perpendicular
to the vector < 4, −2 >.
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Math 171, Benjamin
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Example: Find parametric equations for the line y = 5x − 7.
Determine whether the lines r1 (t) = (−1 − 2t)i + (2 + t)j and r2 (s) = (5 + 3s)i + (3 + 6s)j are parallel,
perpendicular, or neither. If they are not parallel, find the point of intersection.
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