Given vectors how to find third equation

  • Thread starter Thread starter mill
  • Start date Start date
  • Tags Tags
    Vectors
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
mill
Messages
72
Reaction score
0

Homework Statement



If |v|=4, |w|=3 and the angle between and is pi/3, find |2v −w|

Homework Equations


##cosθ=\frac {v dot w} {|v||w|} ##

The Attempt at a Solution


## 6= v dot w ##

This is as far as I got. How would I find the separate values of v and w for the equation?
 
Physics news on Phys.org
mill said:

Homework Statement



If |v|=4, |w|=3 and the angle between and is pi/3, find |2v −w|

Homework Equations


##cosθ=\frac {v dot w} {|v||w|} ##

The Attempt at a Solution


## 6= v dot w ##

This is as far as I got. How would I find the separate values of v and w for the equation?

Very good! How is the absolute value of a vector defined with dot product by itself?
You do not need to know the individual vectors. You need the absolute value of 2v-w.

ehild
 
ehild said:
Very good! How is the absolute value of a vector defined with dot product by itself?
You do not need to know the individual vectors. You need the absolute value of 2v-w.

ehild

u dot u = ##|u|^2## so |u|= ##\sqrt {u dot u} ## ?

Possibly, that becomes |2v-w|=sqrt(something?) or would |?|(1/2)=|2v-w|

I'm afraid I don't see how things are connecting to the third equation though.

1/2 = (something analogous to u dot v)/|2v-w|?
 
Last edited:
mill said:
u dot u = ##|u|^2## so |u|= ##\sqrt {u dot u} ## ?
Start from that definition. What then is ##\sqrt{(2v-w) \cdot (2v-w)}##?