Original Question: What are the angles between the negative direction of the y a

Question

Original Question:

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

Cramster Answer:

Since the 130° is measured counterclockwise from the +x axis, one may write the vectors as
A = 8.00(cos130°i + sin130°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 · -j/A) = arccos(-6.13/v(-5.142 + 6.132) )
= arccos(-6.13/8.00) = 140°.

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

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) = 99.1°


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 for this problem is correct because the book has the answer. I would think to find the direction of A × (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!

Explanation / Answer

A = -5.14i + 6.13j + 0k B = -7.72i -9.20j B + 3.00k = -7.72i -9.20j +3.00k C = A x (B + 3.00k) = 18.39i + 15.42j + 94.61k The magnitude of the vector is C = sqrt(18.39^2 + 15.42^2 + 94.61^2) = 97.61 The angle between -j and the vector is t C dot -j = |C|*|-j|*cos(t) -15.42 = 97.61 * 1 * cos(t) t = arccos(-15.42/97.6) = 99.1°

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