Respuesta :
Answer:
given,
R = 2i + j+3k
a) magnitude in x = 2
y = 1
z = 3
b) magnitude of R
[tex]R = \sqrt{2^2+1^2+3^2}[/tex]
R = 3.74 units
c) angle between the R and the x, y, and z axes.
[tex]cos \theta_x=\dfrac{2}{3.74}[/tex]
θ x = 57.72°
[tex]cos \theta_y=\dfrac{1}{3.74}[/tex]
θ y = 74.51°
[tex]cos \theta_z=\dfrac{3}{3.74}[/tex]
θ z = 36.68°
For the given vector we have:
a) The magnitudes of the components are:
- x = 2
- y = 1
- z = 3
b) |R| = √14
c) x-axis: 26.56°
y-axis: 63.4°
z-axis: 33.7°
How to work with vectors?
Remember that:
- versor i projects on the x-axis.
- versor j projects on the y-axis.
- versor k projects on the z-axis.
So we can write:
R = (2, 1, 3)
a)
The magnitudes of the components are:
- x = 2
- y = 1
- z = 3
b) The magnitude of R is:
|R| = √( 2^2 + 1^2 + 3^2) = √14
c) The angle between R and x-axis is given by:
Atan(y-component/x-component) = Atan(1/2) = 26.56°
Between the R and y-axis is:
Atan(x-component/y-component) = Atan(2/1) = 63.4°
Between R and the z-axis is:
Atan(x-component/z-component) = Atan(2/3) = 33.7°
If you want to learn more about vectors, you can read:
https://brainly.com/question/3184914