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
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