Given two vectors 
and 
.
(a) Determine the angle 
between 
and 
.
(b) Determine the angle that the vector makes with the 
-axis, and the angle that the vector makes on the 
-plane when it is projected on it.
(c) The vector 
is the projection of vector on the 
-plane. Determine the unit vector along the direction of 
.
(d) Express the vector in terms of cylindrical coordinates, i.e., 
, with vectors 
, 
, and 
defined by the vector . Determine the unit vectors , , and in terms of 
, 
, and , and the values of the coefficients 
, 
, and 
.
Solution.
(a) By the identity 
where is the angle between vectors and ,
![\begin{aligned} \cos\theta & = \frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}||\mathbf{B}|} \\ & = \frac{(-16)}{(\sqrt{21})(\sqrt{38})} \\ & = \frac{-16}{\sqrt{798}} \\ \theta & = \arccos \bigg( \frac{-16}{\sqrt{798}} \bigg) \\ & \Bigg[ \textrm{\scriptsize{OR}} \quad \cos^{-1}\bigg( \frac{-16}{\sqrt{798}} \bigg) \Bigg] \end{aligned}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-f78fea4161372f62dbf5880f9704698f_l3.svg "Rendered by QuickLaTeX.com")
(b) Let the -axis be denoted by the vector 
.
Then
The angle between vector and -axis 
is given by
where the sign function 
is defined below
Let the -plane be denoted by the plane of vectors 
for any 
.
Then,
Hold on, I smell a rat. Let’s go another way round. Find the angle 
between vector and the 
-axis 
.
As a matter of routine, take dot product on and 
.
The angle that the vector makes on the -plane is thus 
.
Back to the original line of thought. I.e.,
When the vector is projected onto the -plane (i.e., 
), the angle of projection is defined the angle between and its orthogonal projection 
on that plane. It is clear that 
.
The more careful should have I written
As 
, or, 
.
One should check that
.
By brute force
(c) 
Thus,
(d)
As
.