Website Title: Calculating Angles Between Vectors in Three Dimensions

[enryiggins]

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?

 

[mathman]

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).

 

[tiny-tim]

Welcome to PF!

oh-oh-oh-enryiggins! Welcome to PF!

I am trying to teach myself AS maths.

Just you wait, enryiggins, just you wait.
You'll be sorry, but your tears will be too late. ​

… 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?

 

[enryiggins]

Many thanks for your help. My drawing was not correct hence the confusion.