Parallelogram law of vector addition

Click For Summary

Discussion Overview

The discussion revolves around the parallelogram law of vector addition, focusing on the relationships between vectors, angles, and triangle congruency in a geometric context. Participants explore the implications of these relationships and question the correctness of certain assumptions and calculations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant claims that according to a diagram, vector P equals vector BC and vector Q equals vector OB, leading to the conclusion that angle alpha equals theta/2, but expresses uncertainty about this conclusion.
  • Another participant suggests drawing a specific diagram with P=Q and theta=60 degrees to verify the earlier claims.
  • A participant notes the similarity of triangles OAC and OBC, questioning the relationship between the angles when the magnitudes of OA and OB are equal.
  • There is a correction regarding the assumption that angles AOC and BOC are equal due to triangle congruency, clarifying that it is actually angles ACO and BOC, as well as AOC and BCO, that are equal.
  • A later reply acknowledges a misunderstanding about triangle congruency but reiterates the correction regarding the angles.

Areas of Agreement / Disagreement

Participants express differing views on the relationships between angles and the implications of triangle congruency, indicating that multiple competing interpretations remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the angles and the conditions under which the triangles are considered congruent. The discussion does not resolve the mathematical steps involved in verifying the claims.

parshyaa
Messages
307
Reaction score
19
download4.jpg

  • Acording to this diagram vector P = vector BC and vector Q = vector OB(their magnitudes are also respectively equal.)
  • Therefore acoording to the congruency of triangles angle alpha = (theta)/2. But this is not right( what's wrong )
  • {Tan(alpha) = Qsin(theta)/(P + Qcos(thets)) ]
  • Why resultant vector of P vector and Q vector touches C to form OC vectors.
  • Does the length of vector represents its magnitude.
 
Physics news on Phys.org
parshyaa said:
Therefore acoording to the congruency of triangles angle alpha = (theta)/2. But this is not right( what's wrong )

you have found something wrong...
so why not draw a good diagram say P=Q and angle theta=60 degree and check !
 
  • Like
Likes   Reactions: Simon Bridge
You are observing that OAC and OBC are similar (congruent) triangles.
If |OA|=|OB| (ie P and Q have the same magnitudes) then you have similar isosceles triangles.
What is the relationship between the angles then?
 
You seem to be confused into thinking that angles AOC and angle BOC are equal because the two triangles are congruent.But its actually the angles ACO and BOC;AOC and BCO that are equal.
 
Ohh sorry . It was just basic congurency .
Daymare said:
You seem to be confused into thinking that angles AOC and angle BOC are equal because the two triangles are congruent.But its actually the angles ACO and BOC;AOC and BCO that are equal.
,
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
8K
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
17K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
6K