Question

Please give the answer in detail

Answer 1

There are two ways to solve this question :



1)  Let the angle between P and q is ∅
now, P + q = -r take magnitude 
p² + q² +2p.qcos∅ = r²P² +P² +2p.qcos∅= 2P² cos∅ = 0∅ =π/2 
angle between q and r ∅' q + r = -P
take magnitude 
q² + r² +2rpcos∅'=p² 
p² + 2p² +2.√2P.P cos∅' =P² cos∅' =-1/√2 ∅ = 3π/4 
angle between r and P is ∅"P + r = -q 
take magnitude 
P²+ 2P² +2√2p²cos∅" = P² 
cos∅" =-1/√2 
∅" = 3π/4 



                                                            OR


2)Vectors   p + q + r = Ogiven  |p|  = | q|        and   |r | =  √2 | p | Ф = angle between vectors  p and q.
So vector addition  p + q =  - r           |  p + q |² =  | - r |² = | r |²           | p |² + | q |²  + 2 | p | *|q| * CosФ =  2 | p |²           Hence,   cos Ф = 0  =>  Ф = π/2
As p and q vectors of equal magnitude,  - r  will be at  π/4 angle from either p and q.  So r will be at 180 deg from - r.
Angle between  q and r =  180 -45 = 135 deg Angle between   vectors r and  p =    135 deg.
We can also find these above angles by  using vector addition formula as given above.