Solution:-

p+q = r or p-r = -q

(p-r).(p-r) = (-q).(-q)

p2+r2-2prcosθ1=q2

cosθ1 = 1/2 (as p=q=r)

θ1 = 60degree

when p+q+r = 0 or p+r = -q

(p+r).(p+r) = (-q)(-q)

p2+r2+2prcosθ2 = q2

cosθ2 = -1/2 (as p=q=r)

θ2 = 120 degree

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Hello aspirant,

Given   P+Q=R and P=Q=R

When P+Q=R or P - R = - Q

(P - R) (P - R) = (- Q)(- Q)

P2-R2-2PR cos o1°(teta)=Q2

Coso1° = 1/2(as P=Q=R)

o1°=60°

Therefore,angle between Pand Q is 60°

I think my suggestions and knowledge may be helpful to us to u.

Thank you!!!

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