How Do You Calculate Vector Components and Angles in 3D Space?

  • Thread starter Thread starter Neophyte
  • Start date Start date
  • Tags Tags
    Magnitude Vector
AI Thread Summary
The discussion focuses on calculating vector components and angles in 3D space using the vectors a, b, and c. The x, y, and z components of the resultant vector r are determined to be 21, 7, and -2.2, respectively. The angle between r and the positive z-axis is calculated using trigonometric relationships, yielding an angle of approximately 84.3 degrees. The participants express confusion regarding the component of vector a along the direction of vector b and the magnitude of the component of a that is perpendicular to b but lies in the plane of a and b. Clarifications on using dot products and cross products for these calculations are discussed, emphasizing the importance of understanding vector relationships in 3D space.
Neophyte
Messages
42
Reaction score
0

Homework Statement



What are
(a) the x component,
(b) the y component, and
(c) the z component of
r= a-b +c if
a = 6.5i + 4.2j - 6.7k,
b = -6.3i + 1.5j + 1.4k, and
c = 8.2i + 4.3j + 5.9k.
(d) Calculate the angle between r and the positive z axis.
(e) What is the component of a along the direction of b?
(f) What is the magnitude of the component of a perpendicular to the direction of b but in the plane of a and b and ?


Homework Equations


a . b = abcosθ
c = absinθ
tanθ= (aj/ai)

The Attempt at a Solution


a) 21
b) 7
c) -2.2

21i + 7j - 2.2k

d)√(441+ 49 + 4.84) = 22.24

θ=sin-1 = 7/22.24
θ = 18.34 /*Incorrect I am pretty sure ;/ */
I took y/r r as hypot and y as opposite

e)Well I know how to find the ay = asinθ etc but the z I do not know.
So I tried
direction of b -i,+j,+k
thus a = -6.5i + 4.2j + 6.7k but then I believe I have three components. So I am clueless.
f) close the book
a X b produces third vector c

i j k
6.5i + 4.2j - 6.7k
-6.3i +1.5j + 1.4k

= 15.93i - 33.11j + 36.21k
= magnitude 51.58
 
Physics news on Phys.org


d) You have to find the angle between -z-axis and r, calculate cosθ = z/r, then find (180 - θ ) to find the angle with + z-axis.
 


rl.bhat said:
d) You have to find the angle between -z-axis and r, calculate cosθ = z/r, then find (180 - θ ) to find the angle with + z-axis.

so cosθ = 2.2/22.24
θ = 84.3
180 - θ = 95.6 degrees?

If - 2.2k was positive wouldn't it be the same thing as the + z-axis because cos is same for +/-? Will I always do (180 - θ)?

I had done e) wrong
(e) What is the component of a along the direction of b?
Its basically
a.b = abcosθ
a.b / b = component along direction of b

(f) What is the magnitude of the component of a perpendicular to the direction of b but in the plane of a and b and ?
What is it talking about :cry:

I thought it had something to with
aXb= c = absinθ

But then c is the direction perpendicular to the plan of a and b not in it.
So is a = a; b = c; c=a;
aXc = b = acsinθ
But I don't have the c :\\\\\\\\\\\\
:confused:
 


e) component of a along b is acosθ = abcosθ/b = a.b/b
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top