I'm asking for clarification on following question :
Find the direction Cosines of AB and hence calculate the angle in degrees ,between AB and each of the positive coordinate axes.
AB = = -5i,13j,-0.5k or -5x,13y,-0.5z
My confusion lies with 'and each of the positive coordinate axes.'
I have calculated a direction cosine using:
A.B = (5*0)+(13*0)+(5*4.5) = 22.5 (2d.p)
|A| = (5^2+0^2+5^2) = root(25+25) = 7.07 (2d.p)
|B| = (0^2+13^2+4.5^2) = root(169+20.25) =13.76 (2d.p)
Cos^-1(22.5/7.07*13.76) = 76.60 degrees (1d.p)
Perhaps this next part is asking me to work out the angle from each 0x,0Y.0Z points respectively, but it's unclear to me.
Many thanks for any clarification/advice regarding.
It has been difficult at first to understand that the coordinates of $A$, resp. $B$ are $(5,13,5)$ and $(0,0,4.5)$, and to understand after that that we don't need them...
If the question is about vector $\vec{AB}$, you can forget $A$ and $B$. So the first formula you use $A.B=...$ is for computation of another angle (angle $AOB$).
A second thing : the x-axis is directed by vector $\vec{u}(1,0,0)$. Thus, if you want the angle between $\vec{AB}$ and $\vec{u}$, you apply "your" formula in the following way:
$\vec{AB}.\vec{u}=\|\vec{AB}\|\|\vec{u}\|\cos(\alpha)$ or
$(-5*1)+(13*0)+(-0.5*0) = \sqrt{50}\sqrt{1} \cos(\alpha)$
(note that I have kept the minus signs : there is no reason for changing the minus signs into plus signs).
Then $\cos(\alpha)=-5/\sqrt{50}=-1/\sqrt{2}$. Thus $\alpha=3 \pi/4$, i.e., 135 degrees.
Same type of computation for the other angles.
Final remark: forget the term "direction cosines" which is now obsolete and may be confusing.