Chapter 6 Vectors6.1 General Information and Geometric uses

A vector is a quantity that has both magnitude and direction.

tip

tail

Two vectors are equivalent iff they have both the same magnitude and same direction.

equivalentnot equivalent

not equivalent

Notation: a vector is typically shown one of 2 ways. In typed text, it is shown as a lower case letter that is bold such as u, v, a, b, etc.when handwritten, draw an arrow over the letter. I will draw a half arrow-head like this:

Vector addition: Scalar multiplication:

Zero vector

Additive inverse

Linear combination

Magnitude (norm) and properties

(triangle inequality)

00

viffv

vrvr

vuvu

EX 1 Given vectors u and v as shown, finda) 2u + vb) u – vc) 3u – ½v

uv

EX 2 Express u as a linear combination of v and w

u v

1/3 w 1/3 w 1/3 w

EX 3 a) Express u and v in terms of a and bb) Express a and b in terms of u and v

u v

a

b

Additional notation conventions

v1. can also mean magnitude, depending on the source.

Example usage: find the magnitudes of the horizontal and vertical components of the vector, v, with = 10 and a direction, θ = 60°v

2. Vectors can be written in a way clearly showing its horizontal and vertical components:

Example usage: find the magnitude and direction of vector, v,

vertical,horizontal

8,15

6.2 Vectors and NavigationDefinitions (look in text for exact words; this slide shows examples:

BearingHeadingTrue courseGround speedAir speed

N, 0°

E, 90°

S, 180°

W, 270°

EX 5 A plane flies due west for 400 km and then on a heading of 290° for 250 km in one hour. Find the plane’s true course and ground speed.

EX 5 A boat leaves the dock and travels to a lighthouse 30 km away with a bearing of 65°. It then travels to an island 100 km away, traveling with a bearing of 290°. If the boat wanted to travel directly from the dock to the island, how far would it travel and what would be the correct bearing to take?

EX 5 A boat leaves the dock and travels to a lighthouse 30 km away with a bearing of 65°. It then travels to an island 100 km away, traveling with a bearing of 290°. If the boat wanted to travel directly from the dock to the island, how far would it travel and what would be the correct bearing to take?

EX 5 A boat leaves the dock and travels to a lighthouse 30 km away with a bearing of 65°. It then travels to an island 100 km away, traveling with a bearing of 290°. If the boat wanted to travel directly from the dock to the island, how far would it travel and what would be the correct bearing to take?

One last addition to this problem (to emphasize bearing): What is the bearing from the island to the dock?

When the magnitude of a vector represents force and the object is at equilibrium (at rest or no acceleration), we will sum all of the vector components to zero.

6.3 Vectors and Force

a

u v u

a

In this section it will be helpful to have parallelograms that are rectangular (above) or a triangle of vectors that sum to the zero vector 0 (below).

v

θv cos θ

v sin θ

c

edc + d + e = 0

Recall that the sum of two vectors will result in the diagonal of a parallelogram.

EX 6 A 50 kg box slides with constant velocity down a ramp inclined at 15° with the horizontal. Find the frictional force acting on the box.

Helpful facts: in metric measure, force has the units and we give it the symbol N for Newton.

In metric measure, the resultant force when gravity acts on any object can be found by force = (mass)(acceleration) where a = 9.8 m/s2

f

g

15°

2s

mkg

f is the component of the gravitational force parallel to the surface and is the force that must counteract the friction we are interested in. The forces parallel to the ramp must balance each other out: f + (-f) = 0)

EX 7 A weight is suspended from two ropes, as shown. The weight has a mass of 50 kg. Find the tension in each rope.

35°

50°

50 kg

EX 8 The given forces act on a particle, P. Find the magnitude and direction of a force F that will keep P stationary.

F1: 30 N; 35°F2: 25 N; 130°F3: 10 N; 290°

In this section we use another convention for naming vectors.

Let i be a vector with a magnitude of 1 having a horizontal direction and j be a vector with a magnitude 1 having a vertical direction. Then we can write any vector as a linear combination of its horizontal and vertical components.

6.4 Algebraic Vectors and Unit Vectors

if w = 3i – 6j we travel to the right 3 and down 6 (3 times a positive i and 6 times a negative j)

if v = ai + bj then the magnitude of v is

22 ba

if u = ai + bj and w = ci + dj thenu + w = (a + c)i + (b + d)j(add the horizontal components and add the vertical components)

If u = 2i + j and w = 3i – 6j

Find u + w

Find u – w

Unit Vectors

A unit vector is defined to be a vector with a magnitude of 1 in the same direction as the given vector. Example: find a unit vector parallel to v if v = 3i + 4j.

To find the unit vector of v we divide v by its magnitude.

Let u be the unit vector in the direction of v:

6.5 The dot product, angle between two vectors, orthogonal vectors

The only product regarding vectors that we have considered so far is the scalar product . We consider the product of two vectors – the dot product.

But, let us first consider the angle between two vectors:

If u = ai + bj and v = ci + dj and if Ѳ is the angle between the two vectors then:

vucos

bdac

v

u

v – u

Ѳ

Proof:Using the law of cosines |v – u |2 = |u|2 + |v|2 – 2 |u||v| cos ѲSince v – u = (c – a)i + (d – b)j we can write(c – a)2 + (d – b)2 = a2 + b2 + c2 + d2 – 2 |u||v| cos Ѳc2 – 2ac + a2 + d2 – 2bd + b2 = a2 + b2 + c2 + d2 – 2 |u||v| cos Ѳc2 – 2ac + a2 + d2 – 2bd + b2 = a2 + b2 + c2 + d2 – 2 |u||v| cos Ѳ– 2ac – 2bd = – 2 |u||v| cos Ѳac + bd = |u||v| cos Ѳ solving for cos Ѳ we get:

vucos

bdac

EX 9Find the angle between u = 3i – 4j and v = 5i + 12j

vucos

bdac

We define the dot product of two vectors u = ai + bj and v = ci + dj to be:

bdac

bdac

vu

vuvuvu

cosvuvu

vucos

bdac Note also that our equation becomes

vu

vucos

Going back to matrix analogies, recall that the product of two matrices a NOT a matrix, but a single number. The same is true about the multiplication of two vectors – the result is a number. We call the multiplication of two vectors the “dot product”.

EX 10 Find the dot product of u = 2i – 3j and v = i + 2j

Another important property: the vectors u and v are parallel to each other iff their dot product = ±|u||v|

180isthembetweenanglethewhen

0isthembetweenanglethewhen

vuvu

vuvu

An important property: the vectors u and v are orthogonal to each other (perpendicular to each other) iff their dot product is 0.

Some practice on scratch paper :

For the given vectors write a vector orthogonal to it. Helpful hint – think of slopes of perpendicular lines

a. 3i + 2jb. 6i – 4j c. -4/5 i + 2/5 j

Some answers:a. -2i + 3j, 2i – 3j, -4i + 6j

b. -4i – 6j, -2i – 3j, 4i + 6j c. 2/5 i + 4/5 j, 2i + 4j, 3i + 2j

a. 3i + 5jb. -2i + 4j

Some answers:a. -3i – 5j, 6i + 10j, 3/2 i + 5/2 j

b. -i + 3j, 2i – 4j, any fractional ones as well

For the given vectors write TWO vectors parallel to it. Helpful hint – think of slopes of parallel lines

Suppose the vectors v and a are represented by the figures shown below. Then the vector projection of v onto a is the vector va as shown.

v

a

va

To find the value of the vector projection we go back to the example where we first found a unit vector, we determined that a unit vector parallel to w if w = 3i + 4j was u = 3/5 i+ 4/5 j.

w = 5 uw = 5 uw = quantity of unit

vectors needed for the magnitude of w

Unit vector in the correct direction

va =quantity of unit vectors needed for the magnitude of va

Unit vector in the correct direction

EX 11a Find the vector projection of v onto a, va where v = -i + j and a = 3i + 4j

The scalar projection is the magnitude (norm) of the vector projection. We will compute the value of the scalar projection as follows:

the dot product of (the unit vector in the direction of a) and (v).

We can think of the scalar projection as the distance along the vector that the other one would “cast a shadow over” if the sun were perpendicular to the vector being projected onto.

v

ava

We can think of the scalar projection as the distance along the vector that the other one would “cast a shadow over” if the sun were perpendicular to the vector being projected onto.

v

ava

EX 11b Find the scalar projection of v onto a, |va| where v = -i + j and a = 3i + 4j

va

w

v

a

va

Note that the same process for finding the value of the scalar projection would work no matter how the vectors “looked” relative to each other:

Our example from the previous slides gave us |va| < |a|

This example would give us |va| > |a|

v

a

va

This example would give us |va| < |a|, but we will also be interested in the vector projection (next slide) which would have a direction negative to a.

In general we will be interested in 3 things:

1. Find the scalar projection of v onto a (a numerical value)2.Find the vector projection of v onto a (a vector)3.Resolve v into components parallel to two given (orthogonal) vectors

EX 12Resolve v into components parallel to a and to b

v = 2i + ja = 3i – 2jb = 2i + 3j

We have already:1. Found the scalar projection of one vector onto another vector (a numerical value)2.Found the vector projection of one vector onto another vector (a vector)Last, we will:3. Resolve a vector into components parallel to two given orthogonal vectors.

bbb

vbaaa

vav

vvv ba

For part 3 above, we put together two pieces from the last idea. If a and b are orthogonal:

EX 13aFind just vb:

On the previous slide we resolved v into components parallel to two given orthogonal vectors, a and b. We used the formulas below. But what does that really mean?

bbb

vbaaa

vav

vvv ba

v = 2i + ja = 3i – 2jb = 2i + 3j

EX 13bFind just |va|: