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If the vectors 3p+q;5p-3q and 2p+q;4p-2p are pairs of mutually perpendicular vectors then sin(p^q) is [here i have written p and q for p bar and q bar,those are vectors,sorry for inconvience] (A)√55/4 (B)√55/8 (C)3/16 (D)√247/16 sin(p^q) means sin of angle b/w p and q please give detailed solution please help

If the vectors 3p+q;5p-3q and 2p+q;4p-2p are pairs of mutually perpendicular vectors then sin(p^q) is [here i have written p and q for p bar and q bar,those are vectors,sorry for inconvience]


(A)√55/4 (B)√55/8 (C)3/16 (D)√247/16


sin(p^q) means sin of angle b/w p and q


 


please give detailed solution please help

Grade:12th Pass

4 Answers

Aman Bansal
592 Points
12 years ago

Dear Suchita,

(3p+q)x(5p-3q) = -9pxq+5qxp = 0 thus p x q = 0

Thus solve this equation to get angle between the two.

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suchita undare
20 Points
12 years ago

by mistake i have written 2p+q;4p-2p it should be 4p-2q so sorry

vasanth
11 Points
6 years ago
sorry brothers !... 
   what ever u have said is worry one iam really sorry for it........
the ans is 
   3p+q and 5p-3q are perpendicular . therefore ,
           (3p+q).(5p-3q)=0
       15p^2 -3q^2=4p.q    (i)
2p+q and 4p-2q are perpendicular .therefore,
            (2p+q).(4p-2q)=0
          8p^2=2q^2
        q^2=4p^2          (ii)
 now,       
           cos(-) = p.q/|p||q|
   substituting q^2=4p^2 in (i),3p^2=4p.q
  cos(-) = ¾ .p^2/|p|2|p| = 3/8
 
or (-) =       cos-13/8 
 and this is the ans 
 thank you …..
sooo this was my own try and i just confused by seeing ur ans and iam so sorry to say this 
if u dont no the ans plz dont missuse the net what u have becz we are not that much to get it 
 but i think that u r not like that and plz if u knew that correctly plz upload it 
  so plz dont confuse anybody and make them to feel bad abt u 
 thank u but brcz of u i have learnt many things and thank u for that …....
saksham
15 Points
5 years ago
Ans B
First take dot product of both vectors and equal them to 0
Second from first two vectors and second two vectors generate the value of
cos theta
Last use sin theta =sqrt(1-cos^2 theta)
 
 
 
hope you got the ans

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