Q 21 solution â ‹Q 21 The vectors AB→ = 3i^ - 2j^ +2k^ - Maths - Vector Algebra

Question

Q 21 solution

â€‹ Q   21       T h e  
v e c t o r s   A B →   =   3 i ^   -
  2 j ^   + 2 k ^   a n d   B C →   =
  - i ^   -   2 k ^     a r e   t h e
  a d j a c e n t   s i d e s   o f   a  
p a r a l l e log r a m   T h e   a n g l e   b e t
w e e n   i t s   d i a g o n a l s   i s   A
      π 4          
                  B
        π 3        
                 
  C           3 π 4    
                 
  D           2 π 3

Ankita Agarwal · Accepted Answer

Dear Student,

The two diagonals are given by AB-BC = 3i^-2j^+2k^--i^-2k^  = 4i^-2j^+4k^and AB+BC = 3i^-2j^+2k^+-i^-2k^ = 2i^-2j^AB-BC  = 42+-22+42AB-BC  =16+4+16AB-BC  = 6Similarly, AB+BC = 22+-22AB+BC = 4+4AB+BC = 8AB+BC = 22Taking scalar product, we have, cos θ = 4i^-2j^+4k^
2i^-2j^+0k^4i^-2j^+4k^2i^-2j^cos θ = 8+46×22 cos θ = 12122These vectors have magnitude 6 and 22, respectively, and their dot product is 12
Therefore the angle between them is cos-112622 = cos-1 12 = π4Hence, option A is correct

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Q . 21 solution ​ Q . 21 T h e v e c t o r s A B → = 3 i ^ - 2 j ^ + 2 k ^ a n d B C → = - i ^ - 2 k ^ a r e t h e a d j a c e n t s i d e s o f a p a r a l l e log r a m . T h e a n g l e b e t w e e n i t s d i a g o n a l s i s A π 4 B π 3 C 3 π 4 D 2 π 3