Unit 5 - Activity 10

ACTIVITY 5.10
THE THEOREM OF PYTHAGORAS FOUNDED AGAIN

If in the previous activity we have showed a theorem of Thales using scalar products, in this proof we will show the Pythagoras theorem, in a direct way and in an opposite way, using also scalar products.

Theorem of Pythagoras in direct way:

Being a ABC right triangle in A, and being

and

, that is to say, the cathetus (see in the figure). Then we can put the hypotenuse

like the difference

; we calculate its module to the square:
|

|² = |

|² = (

)² =

² - 2

² =

²+

² = |

|²+ |

|²
because to be two perpendicular vectors

and

, its scalar product is zero. We get then that the square of the hypotenuse is the sum of the squared cathetus.

Theorem of Pythagoras in opposite way:

Being a ABC triangle that verifies ||² = ||²+ ||², (being =, = and ==-). Then, like always is verified
||² = |-|² = (-)² =² - 2·+² = ||² - 2·+ ||²
has to be necessarily 2·, therefore ·= 0, the two vectors and are perpendicular, and the ABC triangle is rectangle with a right angle in A.

INTERACTIVE ACTIVITY

This applet shows that if two vectors and are perpendicular (that is to say, ·= 0) then ||,
|| and |-| verifies the relation:
|-|² = ||²+ ||²(that we will call relation of Pythagoras) and reciprocally.

Which of next pairs of vectors and do they verify the relation of Pythagoras
|-|² = ||²+ ||² ?

1)=(2,5) and=(7,-3)
2)=(-2,5) and=(10,4)
3)=(2,4) and=(6,-2)
4)=(6,9) and=(3,-2)

SOLUTION

HOMEWORK

(Do you want a Word document prepared for making this work ?)

Show the theorem of the cosine (a² = b²+c²-2bcCosA) using scalars products.

Instruction: with the same used notation in the theorem of Pythagoras (that is to say, =, = and ==-), calculate ||² = |-|² , and you take into account that if · isn't zero.

END OF ACTIVITY 5.10
THE THEOREM OF PYTHAGORAS FOUNDED AGAIN

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