The Dot Product's Geometric Relationship
Vector Products Franz Helfenstein NAME
Write answers here! Use extra paper for your work as necessary.
|The Dot Product's Geometric Relationship (Dot Product = Scalar Product) |[pic] |
|Note: Vectors are usually denoted by either or A. That is, = A! |figure 1 |
|1) Use u, v & w and vector addition to write a vector equation for this diagram. | |
| | |
| | |
| | |
|2) Solve for w. | |
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. |[pic] |
| |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. | |
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, y2 and 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 ( v then 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: [pic].
Unfortunately, a 3 × 3 matrix gets significantly more complicated.
The definition of the determinant of a 3 × 3 matrix: [pic]
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 |[pic] |
|dividing by its magnitude. |A Unit Vector |
| | |
|= = | |
|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. | |
|There are 3 special unit vectors; the vectors parallel to the axes. The unit vector parallel to the x-axis is |[pic] |
|denoted by i (or ). Similarly, j (or ) denotes the unit vector parallel to the y-axis and , k (or ) denotes the |The Standard Basis Vectors |
|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 unit vectors i, j and k are called the standard basis vectors. Any vector can also be written in terms of the standard basis vectors in the following manner. u = (ux, uy, uz) = ux i + uy j + uz k.
Example u = (-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 |[pic] |
|A unit vector has the special property that its components are given by |[pic] |
| |The Direction Cosines |
|= (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 cross product can always be calculated by the determinant. u × v = [pic] where i, j and k are basis vectors. For u, v in the plane, the computation simplifies to.
u × v = [pic]
The cross product is not commutative. u × v ≠ v × u
The Cross Product's Geometric Relationship
| |[pic] |
|1) Show that u ( w & v ( w by showing u · w = 0 & v · 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 |[pic] |p=(2,3) q=(1,–4) r=(5,1) |[pic] |
| | | | |
|Find u = (x,y). | |Write s in terms of p, q and r | |
|Find the length of u. | |Find s = (x,y). Find ||s||. | |
|Find its reference angle θ. | |Find its reference angle θ. | |
|p=(2,3) q=(3,4) r=(5,–1) |[pic] |v = (x,y) |[pic] |
| | |Find || v || | |
|Write s in terms of p, q and r | |Find θ , Find φ | |
|Find s = (x,y). Find ||s||. | | | |
|Find its reference angle θ. | | | |
| |[pic] |F1 = 140 lbs N20°W |[pic] |
|The lever is 18" long. The force F is | |F2 = 120 lbs N 20° E | |
|60 lbs. | |F3 = 125 lbs S 30° W | |
| | |F4 = 150 lbs S 30° E | |
|Find the torque, τ | |Find the resultant force | |
| |[pic] | |[pic] |
|Find the minimum force necessary to | |Find the force, F, needed to obtain the| |
|push the cart up the ramp. | |position shown. | |
| |Find force A and force B. |Find force A and force B. |
|Find force A and force B. | |[pic] |
| |[pic] | |
|[pic] | | |
|Find the total upward force on a (20' × 30') roof|The river flows at 8 mph. How fast must the |A = 425 #, B = 500 # Find C, α, θ to make the |
|with a 6/12 pitch when the wind exerts a force of|swimmer swim upstream to go straight across? θ =|tension balance. |
|40 lbs/ ft2 horizontally. |70° |[pic] |
|[pic] | | |
| |[pic] | |
|A river flows at 8 mph. A swimmer swims straight| | |
|across at 2 mph. |A jet flies at 560 mph in calm conditions. |A jet flies at 560 mph in clam conditions. Suppose |
|What angle will result vs. the river bank? |Suppose it flies N 30° E with a 60 mph tail wind |it flies N 30° E with a 60 mph head wind (S 10° E). |
|[pic] |(N 80° E). | |
| | |How fast is the jet actually flying? |
| |How fast is the jet actually flying? | |
|Find the tension in the cable holding up Joe's |The ferry crosses the river by turning at an |Find the tension, T, so that the horizontal forces |
|sign. The sign weighs 230#. The cable is |angle while sliding on a cable. The river exerts|balance. The hill slopes at 30° |
|attached in the middle of the 4 ft wide sign. |a force of 30,000# on the ferry. When it's |[pic] |
|[pic] |turned θ=15° how much force is pushing it across.| |
| |[pic] | |
[pic]
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- dot product matrix matlab
- dot product matrices
- dot product matrix calculator
- knapp s model relationship stages
- when was the dot com bust
- compute the unit product cost
- the end product of glycolysis is
- the end product of glycolysis is quizlet
- dot product in numpy
- dot product calculator
- python dot product numpy
- the new product development process