Use the cross product to find the sin of an angle then prove it

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Homework Help Overview

The problem involves finding the angle between two vectors using both the dot product and the cross product, specifically focusing on calculating the sine of the angle and confirming the Pythagorean identity involving sine and cosine.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for the cross product and the norms of the vectors, questioning the validity of their results and the errors encountered when attempting to find the sine of the angle.

Discussion Status

Some participants have provided insights into the potential reasons for the errors encountered, particularly regarding the ranges of the inverse sine and cosine functions. There is an acknowledgment of differing angle results, and the discussion includes verification of the relationship between sine and cosine values.

Contextual Notes

Participants note the importance of understanding the ranges of trigonometric functions and how they relate to the angles in different quadrants, which may affect the interpretation of results.

shemer77
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Homework Statement


Leta theta be the angle between the vectors u=2i +3j -6k and v=2i + 3j+6k
A) use the dot product to find cos theta
b) use the cross product to find sin theta
c) confirm that sin^2(theta) + cos^2(theta)=1

The Attempt at a Solution


I got a to be 117 degrees, but however b and c are stumping me, whenever I try to do b I get an the cross product and then try to solve for theta but i get an error using my calculator, I rechecked my calculations twice and everything make sense am I doing something wrong or is it really not possible.
 
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It would be more helpful to someone assisting you if you explained what you did for part (b), in order to understand why you are getting an error...

(And I get an angle of 118º for theta.)
 
well all i did was the cross product of u and v which came out to 36i -24j+0k and i took the norm of that which came out to 12*sqrt(13) and I divided that by the norm of u * the norm of v which is 49. so I had 12*sqrt(13)/49 and i took the inverse sin of that. which would obviously give me an error. But I don't see what I did wrong all my calculations are correct.
 
shemer77 said:
well all i did was the cross product of u and v which came out to 36i -24j+0k and i took the norm of that which came out to 12*sqrt(13) and I divided that by the norm of u * the norm of v which is 49. so I had 12*sqrt(13)/49 and i took the inverse sin of that. which would obviously give me an error. But I don't see what I did wrong all my calculations are correct.
Check how you have input \frac{12\sqrt{13}}{49} ; this is a positive number smaller than 1 . Is your "error" that you don't get the same angle as you did for the arccosine? Don't forget that the range of arccosine is 0º to 180º , but the range of arcsine is -90º to 90º ; the arcsine result for theta (out of a calculator) is the supplement of the correct value. (The angle theta is in the second quadrant.)

Part (c) sidesteps this anyway, since you are simply asked to sum the squares of your values from parts (a) and (b)... [EDIT: You must have gotten cos(theta) = -23/49 to get the angle of 118º . The sum of the squares does give you 1 .]
 
Last edited:
ah awesome man, I got it thanks for your help!
 

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