PHYSICS 2 Lecture Notes - Lecture 12: Dot Product, Joule, Euclidean Vector

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PHYSICS 2 Lecture Notes - Lecture 12: Dot Product, Joule, Euclidean Vector

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Document Summary

Physics 2 - lecture 12 - scalar products of vectors and relative velocity. The scalar product is a way of multiplying two vectors that results in a vector. The scalar product is also known as the dot product. The scalar product of two vectors, a . B: the projection of the vector a , the magnitude of vector b. This is useful is you want to know the component of one vector that lies along the direction of another vector. The dot product of two vectors can be negative. Calculating the dot product of vectors in polar form. If you know the magnitude and direction of each vector, you can calculate their dot product using the formula: A||b|cos( are the magnitudes of the vectors a . And b is the angle between the vectors a . If the two vectors point in the same direction ( then the dot product is.

Related Questions

The magnitude of the vector (i.e., cross) product of vectors A andBis twice as large as the magnitude of the scalar (i.e., dot)productof the same two vectors. A and B, if they were placedtail-to-tailin space, would form an angle of approximately

a) 63^o
b) 27^o
c) 207^o
d) 117^o
e) none of the above are even approximately correct

I already know the answer (and yes, there are two possibleanswers),but I need someone to work this out step by step becauseI'm reallyconfused how to figure this out.

Original Question:

What are the angles between the negative direction of the y axisand (a) the direction of A, (b) the direction of the product A Ã B,and (c) the direction of A Ã (B + 3.00k)?

Cramster Answer:

Since the 130Â° is measured counterclockwise from the +x axis, onemay write the vectors as
A = 8.00(cos130Â°i + sin130Â°j) = -5.14i + 6.13j
B = Bxi + Byj = -7.72i -9.20j.

a) The angle between -j and the direction of A is
? = arccos(A Â· -j/A) = arccos(-6.13/v(-5.142 + 6.132) )
= arccos(-6.13/8.00) = 140Â°.

b) Since the y axis is in the xy plane, and AxB is perpendicular tothat plane, the angle is 90Â°

c) The vector can be simplified to
Ax(B + 3.00k) = (-5.14i +6.13j)x(-7.72i -9.20j +3.00k)
= 18.39i + 15.42j + 94.61k
The magnitude of the vector is v( i2 + j2 + k2) = 97.6.
The angle between -j and the vector is
? = arccos(-15.42/97.6) = 99.1Â°

MY QUESTION:

What is the equation used in part C) of this problem?

It seems that it should be sine to me though I know the answer forthis problem is correct because the book has the answer. I wouldthink to find the direction of A Ã (B + 3.00k) you would:

1) Use the cross product including the k.

2) Then find the magnitude.

3) Finally use the equation AxB=absin(theta)

What is the correct equation to use?

Please also include an explanation. Thanks!