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