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Vectors, EXT1 V1 EQ-Bank 26

`OABC`  is a quadrilateral.

`P`, `Q`, `R` and `S` divide each side of the quadrilateral in half as shown below.
  


 

Prove, using vectors, that  `PQRS`  is a parallelogram.  (3 marks)

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`text(See Worked Solution)`

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`text(Consider diagonal)\ \ overset(->)(OB):`

`overset(->)(OB) = overset(->)(OA) + overset(->)(AB) = overset(->)(OC)+ overset(->)(CB)`

`overset(->)(PQ) = overset(->)(PA) + overset(->)(AQ) = 1/2(overset(->)(OA) + overset(->)(AB)) = 1/2overset(->)(OB)`

`overset(->)(SR) = overset(->)(SC) + overset(->)(CR) = 1/2(overset(->)(OC) + overset(->)(CB)) = 1/2overset(->)(OB)`

`:.overset(->)(PQ) = overset(->)(SR)`

 

`text(Consider diagonal)\ \ overset(->)(AC):`

`overset(->)(AC) = overset(->)(AB) + overset(->)(BC) = overset(->)(AO) + overset(->)(OC)`

`overset(->)(QR) = overset(->)(QB) + overset(->)(BR) = 1/2(overset(->)(AB) + overset(->)(BC)) = 1/2overset(->)(AC)`

`overset(->)(PS) = overset(->)(PO) + overset(->)(OS) = 1/2(overset(->)(AO) + overset(->)(OC)) = 1/2overset(->)(AC)`

`:. overset(->)(QR) = overset(->)(PS)`

 

`text(S)text(ince)\ PQRS\ text(has equal opposite sides,)`

`PQRS\ text(is a parallelogram.)`

Vectors, EXT1 V1 EQ-Bank 24

`PQRS`  is a parallelogram, where  `overset(->)(PQ) = underset~a`  and  `vec(PS) = underset~b`
 


 

Prove, using vectors, that the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of the sides.  (3 marks)

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`text(See Worked Solution)`

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`overset(->)(PR) = underset~a + underset~b,\ \ overset(->)(SQ) = underset~a – underset~b`

`overset(->)(PQ) = overset(->)(SR) = underset~a`

`overset(->)(PS) = overset(->)(QR) = underset~b`

`text(Prove)\ \ |underset~a + underset~b|^2 + |underset~a – underset~b|^2 = 2|underset~a|^2 + 2|underset~b|^2`

`|underset~a + underset~b|^2 + |underset~a – underset~b|^2` `= (underset~a + underset~b) · (underset~a + underset~b) + (underset~a – underset~b)(underset~a – underset~b)`
  `= underset~a · underset~a + underset~b · underset~b + 2underset~a · underset~b + underset~a · underset~a + underset~b · underset~b – 2underset~a · underset~b`
  `= |underset~a|^2 + |underset~b|^2 + |underset~a|^2 + |underset~b|^2`
  `= 2|underset~a|^2 + 2|underset~b|^2\ \ …\ text(as required)`