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Interpersonal Deception Theory: Ten Lessons for Negotiators

59 pages
  • exposé
  • cours - matière potentielle : the negotiation
  • cours - matière potentielle : a negotiation
Introduction Believe me -- one of the few universal truths in life is that everybody lies. Lying is not necessarily the product of a malignant heart. Rather, it is simply a component of our human nature. What is meant by the term lie for the purposes of this paper is engaging in intentional deception during interpersonal (i.e., face-to-face) communications. Some deception is culturally acceptable while other forms are not.
  • principles to the field of negotiation theory
  • negotiation context
  • deceptive communication
  • interpersonal deception theory
  • lie detectors
  • course of the negotiation
  • buller
  • deception
  • truth-bias
  • truth bias
  • individuals
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Geometric Algebra
for Computer Science
Answers and hints to selected drills and exercises
Leo Dorst, Daniel Fontijne and Stephen Mann
October 28, 20102 FINAL | October 28, 2010
When we wrote the drills and exercises for Geometric Algebra for Computer
Science, we intended them to be for self-study. As such, we are tempted to release
solutions to all the drills and structural exercises. However, as some instructors
may wish to use these as homework questions, for now we are only releasing the
solutions to all of the drills and to most of the odd numbered structural exercises.
In the future, we plan to release solutions to all of the structural exercises, since
solutions to these questions will likely appear on the internet anyway.Chapter 2
Spanning Oriented Subspaces
2.12 Exercises
2.12.1 Drills
1. Compute the outer products of the following 3-space expressions, giving the
results relative to the basisf1;e ;e ;e ;e ^e ;e ^e ;e ^e ;e ^e ^eg.1 2 3 1 2 2 3 3 1 1 2 3
Show your work.
(a) (e +e )^ (e +e )1 2 1 3
Worked solution:
(e +e )^ (e +e )1 2 1 3
= e ^e +e ^e +e ^e +e ^e1 1 1 3 2 1 2 3
= e ^e +e ^e e ^e1 2 2 3 3 1
(b) (e +e +e )^ (2e )1 2 3 1
Answer: 2e ^e 2e ^e3 1 1 2
(c) (e e )^ (e e )1 2 1 3
Answer: e ^e +e ^e +e ^e2 3 3 1 1 2
(d) (e +e )^ (0:5e + 2e + 3e )1 2 1 2 3
Answer: 3e ^e 3e ^e + 1:5e ^e2 3 3 1 1 2
(e) (e ^e )^ (e +e )1 2 1 3
Answer: e ^e ^e1 2 3
(f) (e +e )^ (e ^e +e ^e )1 2 1 2 2 3
Answer: e ^e ^e1 2 3
2. Given the 2-blade B = e ^ (e e ) that represents a plane, determine if1 2 3
each of the following vectors lies in that plane. Show your work.
(a) e1
Worked solution:
e ^B = e ^ (e ^ (e e ))1 1 1 2 3
= e ^e ^ (e e )1 1 2 3
= 0
Answer: In the plane.
34 FINAL | October 28, 2010
(b) e +e1 2
Answer: Not in the plane.
(c) e +e +e1 2 3
Answer: Not in the plane.
(d) 2e e +e1 2 3
Answer: In the plane.
3. What is the area of the parallelogram spanned by the vectors a = e + 2e1 2
and b = e e (relative to the area of e ^e )?1 2 1 2
Worked solution:
a^b = (e + 2e )^ ( e e )1 2 1 2
= e ^e e ^e 2e ^e 2e ^e1 1 1 2 2 1 2 2
= e ^e + 2e ^e1 2 1 2
= e ^e1 2
Answer: 1
4. Compute the intersection of the non-homogeneous line L with support vector
e and direction vectore , and the lineM with support vectore and direction1 2 2
vector (e +e ), using 2-blades. Does the basis have to be orthonormal?1 2
Answer: Clearly x should be a linear combination of e and e . So set1 2
x =e +e . The demand that x is on L is x^e = e ^e . This yields1 2 2 1 2
e ^e = e ^e , so that = 1. Similarly, the demand that x be on M1 2 1 2
gives = 1. Therefore the solution is x =e + 2e .1 2
5. Compute (2 + 3e )^ (e +e ^e ) using the grade-based de ning equations3 1 2 3
of the outer product.
Worked solution:
3 3XX
h2 + 3ei ^he +e ^ei =3 k 1 2 3 ‘
k=0‘=0
= h2 + 3ei ^he +e ^ei +3 0 1 2 3 1
h2 + 3ei ^he +e ^ei +3 1 1 2 3 1
h2 + 3ei ^he +e ^ei +3 0 1 2 3 2
h2 + 3ei ^he +e ^ei3 1 1 2 3 2
= 2^e + 3e ^e + 2^ (e ^e ) + 3e ^ (e ^e )1 3 1 2 3 3 2 3
= 2e + 3e ^e + 2(e ^e )1 3 1 2 3
2.12.2 Structural Exercises
n1. The outer product was de ned for a vector space R without a metric, but it
2is of course still de ned when we do have a metric space. In R with Euclidean
metric, choose an orthonormal basisfe ;eg in the plane ofa andb such that1 2
e is parallel to a. Write a =e and b = (cose + sine ), where is1 1 1 2
the angle from a to b. Evaluate the outer product. Your result should be:
a^b = sin (e ^e ): (2.14)1 2
What is the geometrical interpretation?
Answer: The outer product evaluation is straightforward. In the answer,FINAL | October 28, 2010 5
e ^e denotes the unit amount of area in thee ^e -plane. The parallelogram1 2 1 2
spanned by a and b has an e -base of , and a e -height of sin(), so the1 2
factor

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