Vector ProductsFranz HelfensteinNAME

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The Dot Product's Geometric Relationship (Dot Product = Scalar Product)
Note: Vectors are usually denoted by either or A. That is, = A!
1)Use u, vw and vector addition to write a vector equation for this diagram.
2)Solve for w. /
figure 1

3)Let u = (x1, y1), v = (x2, y2). Write w in terms of x1, y1, x2, y2

4)Write the Law of Cosines for the triangle shown here in figure 2.
5)Use the notation || u ||, || v || and || w || for the lengths of A, B and C to write the Law of Cosines for the triangle of figure 1. /
figure 2

6)Write || u || in terms of x1, y1 . Similarly, Write || v || & || w ||. Be sure to use definition (3) for w.

7)Now using only x1, y1, x2, y2 substitute out || u ||, || v || & || w || in the Law of Cosines in definition (5). You should obtain an equation with only x1, y1, x2, y2and cos θ.

8)FOIL out || w ||2 and simplify. Recall: u·v = (x1, y1) · (x2, y2) = x1 x2 + y1 y2 by definition of the dot product. Simplify the above equation to get u·v = || u || || v || cos (θ). Note || u ||2 = u·u

9)Show that if uvthen u·v = 0. Thus, u·v = 0 implies uv.

10)Show that if u || v then · = 1. However, u·v = 1 does not imply u || v. Why?

Note: The dot product (scalar product) is useful for vector applications that involve cosine relationships.

The Cross Product

Cross products have a somewhat more complicated numeric structure than dot products. The common way to compute a cross product is to use a determinant. The determinant of a 2 × 2 matrix: .

Unfortunately, a 3 × 3 matrix gets significantly more complicated.

The definition of the determinant of a 3 × 3 matrix:

Unit Vectors

A vector (without units) of magnitude one (length = 1) is called a unit vector. We can scale any vector to a unit vector by dividing by its magnitude.
= = / Creating a Unit Vector
Unit vectors provide a way to indicate direction in a non-biased manner. We use a special notation to indicate a vector is a unit vector. The symbol '^' (caret or hat) is used to denote unit vectors. Thus, denotes the unit vector parallel to u. /
A Unit Vector
There are 3 special unit vectors; the vectors parallel to the axes. The unit vector parallel to the x-axis is denoted by i (or ). Similarly, j (or ) denotes the unit vector parallel to the y-axis and , k (or ) denotes the unit vector parallel to the z-axis. Specifically,
2-D / 3-D
i = (1, 0) / i = (1, 0, 0)
j = (0, 1) / j = (0, 1, 0)
k = (0, 0, 1)
/
The Standard Basis Vectors

The unit vectors i, j and k are called the standardbasis vectors. Any vector can also be written in terms of the standard basis vectors in the following manner. u = (ux, uy, uz) = uxi + uyj + uzk.

Exampleu = (-7, 5), p = 2i + 10j – 6k

(a) Write u in terms of the standard basis vectors / u = -7i + 5j
(c) Write p in component notation / p = (2, 10, -6)
Unit Vectors and Direction Cosines
A unit vector has the special property that its components are given by
= (ux, uy) = (cos θ, sin θ) = (cos θ, cos β) / Direction Cosines
In 3-D, there is no single measurement that gives the vector's direction synonymous to θ for the 2-D case. So, in 3-D, we measure three angles; the angles from each axis. These are noted by α, β and γ. The components of the unit vector correspond exactly to the cosines of these angles.
= (ux, uy, uz) = (cos α, cos β, cos γ) / Direction Cosines
/

The Direction Cosines

The cross product can always be calculated by the determinant. u × v = where i, j and k are basis vectors. For u, v in the plane, the computation simplifies to.

u × v =

The cross product is not commutative. u × v ≠ v × u

The Cross Product's Geometric Relationship

1)Show that uwvw by showing u·w = 0v·w = 0. Explain why that means they are perpendicular. /

2)Let u = (x1, y1, 0), v = (x2, y2, 0). Show || u × v ||2 = || u ||2 || v ||2 − (u·v)2

3)(u·v)2 = || u ||2 || v ||2 cos2 θ Why? Substitute to show || u × v || = || u || || v || | sin θ |

Note: The Cross Product (vector product) is used to facilitate applications involving sine relationships.

Some Vexing Vector Problems Try using Dot and Cross Products where convenient.

v = (5,0) w = (1,5) u = v + w
Find u = (x,y).
Find the length of u.
Find its reference angle θ. / / p=(2,3) q=(1,–4) r=(5,1)
Write s in terms of p, q and r
Find s = (x,y). Find ||s||.
Find its reference angle θ. /
p=(2,3) q=(3,4) r=(5,–1)
Write s in terms of p, q and r
Find s = (x,y). Find ||s||.
Find its reference angle θ. / / v = (x,y)
Find || v ||
Find θ , Find φ /
The lever is 18" long. The force F is 60 lbs.
Find the torque, τ / / F1 = 140 lbs N20°W
F2 = 120 lbs N 20° E
F3 = 125 lbs S 30° W
F4 = 150 lbs S 30° E
Find the resultant force /
Find the minimum force necessary to push the cart up the ramp. / / Find the force, F, needed to obtain the position shown. /
Find force A and force B.
/ Find force A and force B.
/ Find force A and force B.

Find the total upward force on a (20' × 30') roof with a 6/12 pitch when the wind exerts a force of 40 lbs/ ft2horizontally.
/ The river flows at 8 mph. How fast must the swimmer swim upstream to go straight across?  = 70°
/ A = 425 #, B = 500 # Find C, , to make the tension balance.

A river flows at 8 mph. A swimmer swims straight across at 2 mph.
What angle will result vs. the river bank?
/ A jet flies at 560 mph in calm conditions. Suppose it flies N 30° E with a 60 mph tail wind (N 80° E).
How fast is the jet actually flying? / A jet flies at 560 mph in clam conditions. Suppose it flies N 30° E with a 60 mph head wind (S 10° E).
How fast is the jet actually flying?
Find the tension in the cable holding up Joe's sign. The sign weighs 230#. The cable is attached in the middle of the 4 ft wide sign.
/ The ferry crosses the river by turning at an angle while sliding on a cable. The river exerts a force of 30,000# on the ferry. When it's turned =15° how much force is pushing it across.
/ Find the tension, T, so that the horizontal forces balance. The hill slopes at 30°