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.

Answer 1

Let ABC be a right triangle with $\angle$BAC = 90º. Taking A as the origin, let the position vectors of B and C be $\overrightarrow{b}$ and $\overrightarrow{c}$, respectively. Then,

$\overrightarrow{\mathrm{AB}}=\overrightarrow{b}$ and $\overrightarrow{\mathrm{AC}}=\overrightarrow{c}$

Since $\overrightarrow{\mathrm{AB}}\perp \overrightarrow{\mathrm{AC}}$ $\Rightarrow \overrightarrow{b}.\overrightarrow{c}=0$ .....(1)

Now,

${\left|\overrightarrow{\mathrm{AB}}\right|}^2+{\left|\overrightarrow{\mathrm{AC}}\right|}^2={\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2$ .....(2)

Also,

$$\begin{gathered} {\left|\overrightarrow{\mathrm{BC}}\right|}^2={\left|\overrightarrow{c}-\overrightarrow{b}\right|}^2 \\ =\left(\overrightarrow{c}-\overrightarrow{b}\right).\left(\overrightarrow{c}-\overrightarrow{b}\right) \\ ={\left|\overrightarrow{c}\right|}^2-2\overrightarrow{b}.\overrightarrow{c}+{\left|\overrightarrow{b}\right|}^2 \\ ={\left|\overrightarrow{c}\right|}^2+{\left|\overrightarrow{b}\right|}^2.....\left(3\right)\left[\mathrm{Using}\left(1\right)\right] \end{gathered}$$

From (2) and (3), we have

${\left|\overrightarrow{\mathrm{AB}}\right|}^2+{\left|\overrightarrow{\mathrm{AC}}\right|}^2={\left|\overrightarrow{\mathrm{BC}}\right|}^2$