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. | |

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|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 |

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|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 |

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|= = | |

|Creating a Unit Vector | |

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|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 | |

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|i = (1, 0) | |

|i = (1, 0, 0) | |

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|j = (0, 1) | |

|j = (0, 1, 0) | |

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|k = (0, 0, 1) | |

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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 | |

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|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. | |

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|= (ux, uy, uz) = (cos α, cos β, cos γ) | |

|Direction Cosines | |

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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. | |

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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] |

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|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]

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