# I Parallelogram law of vector addition

1. Jul 20, 2016

### parshyaa

• 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( whats 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.

2. Jul 20, 2016

### drvrm

you have found something wrong....
so why not draw a good diagram say P=Q and angle theta=60 degree and check !

3. Jul 20, 2016

### 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?

4. Jul 20, 2016

### Daymare

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.

5. Jul 20, 2016

### parshyaa

Ohh sorry . It was just basic congurency .
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