NDA_Math_2025_Q29

Question:

The position vectors of three points A, B and C are ab  . and c  respectively, where, ca b   (cos )( sin) .22 What is () () ()ab bc ca    () () ()ab bc ca    equal to?

The position vectors of three points A, B and C are ab  . and c  respectively, where, ca b   (cos )( sin) .22 What is () () ()ab bc ca    () () ()ab bc ca    equal to?

Options:

  • a) o 
  • b) 2c 
  • c) 3c 
  • d) unit vector
  • a) o 
  • b) 2c 
  • c) 3c 
  • d) unit vector

correct answer : a)

Explanation:

ab bc ca
  
×() +×() +×()
= ab ba c
  
×() +×() ×
= ab ba ab
  
×() +×() ×+()coss in22θθ
= ab ba ab
  
×() +× () -× ()coss in22θθ
= ab ab ab
  
×() -× () -× ()coss in22θθ
= ab

×() -+() 
1
22coss inθθ = 0

ab bc ca
  
×() +×() +×()
= ab ba c
  
×() +×() ×
= ab ba ab
  
×() +×() ×+()coss in22θθ
= ab ba ab
  
×() +× () -× ()coss in22θθ
= ab ab ab
  
×() -× () -× ()coss in22θθ
= ab

×() -+() 
1
22coss inθθ = 0

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