Website Title: Calculating Angles Between Vectors in Three Dimensions

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The discussion focuses on calculating angles between a given vector b and the positive x-, y-, and z- directions. The user initially attempted to use sine functions to find these angles but discovered that the correct approach involves using cosine functions instead. It was clarified that the components of the normalized vector correspond to the cosines of the angles with respect to the axes. The user acknowledged confusion stemming from an incorrect drawing, which contributed to the misunderstanding. Ultimately, the importance of using cosine for these calculations was emphasized.
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I am trying to teach myself AS maths. This is from Introducing mechanics a txt book by Jefferson and Beadsworth. exercise 2C question 1 part vi)

Question given vector b=6i-3j-2k;
|b| is established as = 7 by calculation

Question
Calculate the angles between b and positive x-, y- and z- directions.

My answer: these are
sine alpha =6/7 = 59 degrees - subtract from 180 to give angle from z axis


Sine beta =2/7 = 16.6 degrees - this is angle from x axis

Sine theta = 3/7 = 25.4 degrees. Need to add 90 degrees to give angle from y axis.

However these are not the correct answers. The correct answers appear to be cosines of the above. This does not make sense to me can you expalin?
 
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Once you normalize the vector to unit length, the coeficients of (i,j,k) are the cosines with respect to the axes (6/7,-3/7,-2/7).
 
Welcome to PF!

oh-oh-oh-enryiggins! Welcome to PF! :smile:
enryiggins said:
I am trying to teach myself AS maths.

Just you wait, enryiggins, just you wait.
You'll be sorry, but your tears :cry: will be too late. :wink:​
… Calculate the angles between b and positive x-, y- and z- directions.

My answer: these are
sine alpha =6/7 = 59 degrees - subtract from 180 to give angle from z axis


Sine beta =2/7 = 16.6 degrees - this is angle from x axis

Sine theta = 3/7 = 25.4 degrees. Need to add 90 degrees to give angle from y axis.

However these are not the correct answers. The correct answers appear to be cosines of the above. This does not make sense to me can you expalin?

The component in a direction is always multiplied by the cos, not the sin.

After all, cos0º = 1, and if the angle is 0º, then the whole of the vector is the component in that direction … so it has to be cos, doesn't it? :wink:
 
Many thanks for your help. My drawing was not correct hence the confusion.
 

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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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