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23 no. Question Solve it Plzzzzzzzzz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23 no. Question 
Solve it Plzzzzzzzzz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

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Grade:11

1 Answers

Vikas TU
14149 Points
5 years ago
Given that P + Q + R = 0
P + Q = – R
 
squaring both sides we get,
|P|^2 + |Q|^2 + 2P.Q = |R|^2
|P|^2 + |P|^2 + 2P.Q = 2|P|^2 (Gven |P| = |Q| and |R| = root2*|P|)
P.Q = 0
Hence the angle between P and Q is 90 degree as their dot product is zero.
 
Angle between Q and R => costhetha = Q.R/(|Q||R|) =>
From P + Q + R = 0
Q + R = – P
|R|^2 + |Q|^2 + 2Q.R = |P|^2
2|P|^2 + |P|^2 + 2Q.R = |P|^2
Q.R = -|P|^2
 
Hence, costhetha =  -|P|^2/(root2|p|^2) => – 1/root2
thetha  = 135 degree
 
Hence a opion is correct.

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