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Question
(Pythagoras's Theorem) Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
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Solution
Let ABC be a right triangle with \[\angle\]BAC = 90º. Taking A as the origin, let the position vectors of B and C be \[\vec{b}\] and \[\vec{c}\] respectively. Then, \[\vec{AB} = \vec{b}\] and \[\vec{AC} = \vec{c}\]
Since \[\vec{AB} \perp \vec{AC}\]
\[\Rightarrow \vec{b} . \vec{c} = 0\] ...........................(1)
Now
\[\left| \vec{AB} \right|^2 + \left| \vec{AC} \right|^2 = \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2\] , ........................(2)
Also,
\[\left| \vec{BC} \right|^2 = \left| \vec{c} - \vec{b} \right|^2 \]
\[ = \left( \vec{c} - \vec{b} \right) . \left( \vec{c} - \vec{b} \right)\]
\[ = \left| \vec{c} \right|^2 - 2 \vec{b} . \vec{c} + \left| \vec{b} \right|^2 \]
\[ = \left| \vec{c} \right|^2 + \left| \vec{b} \right|^2 . . . . . \left( 3 \right) ........................\left[ \text{ Using }] \left( 1 \right) \right]\]
From (2) and (3), we have
\[\left| \vec{AB} \right|^2 + \left| \vec{AC} \right|^2 = \left| \vec{BC} \right|^2\]
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