Answer:
149.54
Step-by-step explanation:
No idea how I got this, but it is correct. I just plugged it into Omni calculator and it gave me that. :p
Answer:
Option B is correct
the degree of rotation is, 
Step-by-step explanation:
A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.
To find the degree of rotation using a standard rotation matrix i.e,

Given the matrix: 
Now, equate the given matrix with standard matrix we have;
= 
On comparing we get;
and
As,we know:

we get;

and

we get;

Therefore, the degree of rotation is, 
Answer:
if 5=70
2=?
cross multiply=28
add 70+28=98
Step-by-step explanation:
Answer:
1/1 I think I hope this helps
Step-by-step explanation:
Answer:
a.
b.
Step-by-step explanation:
We are given that


a.Substitute z=0
...(1)
..(2)
Subtract equation (1) from equation (2)


Substitute y=0 in equation(1)

The point (6,0,0) lie on a line.

Let 



Therefore, the vector 
Line is parallel to vector a' and passing through the point (6,0,0).
The parametric equation is given by

Using the formula
The parametric equation is given by

Angle between two plane

and 

Using the formula



Where
in degree.