# If vector P+vector Q=vector R ;and if P^2+Q^2=R^2 then find the angle between the vectorP and vectorQ?

Jun 15, 2018

$\boldsymbol{P} , \boldsymbol{Q}$ are orthogonal.

(This should be clear from the start - Pythagoreas' Theorem)

#### Explanation:

$\boldsymbol{P} + \boldsymbol{Q} = \boldsymbol{R}$

• $\left(\boldsymbol{P} + \boldsymbol{Q}\right) \cdot \left(\boldsymbol{P} + \boldsymbol{Q}\right)$

$= {P}^{2} + {Q}^{2} + 2 \boldsymbol{P} \cdot \boldsymbol{Q}$

$= {R}^{2} + 2 \boldsymbol{P} \cdot \boldsymbol{Q} q \quad \square$

And because $\left(\boldsymbol{P} + \boldsymbol{Q}\right) = \boldsymbol{R}$

• $\left(\boldsymbol{P} + \boldsymbol{Q}\right) \cdot \left(\boldsymbol{P} + \boldsymbol{Q}\right) = \boldsymbol{R} \cdot \boldsymbol{R} = {R}^{2} q \quad \triangle$

$\square = \triangle \implies \boldsymbol{P} \cdot \boldsymbol{Q} = 0$

$\boldsymbol{P} \cdot \boldsymbol{Q} = \left\mid \boldsymbol{P} \right\mid \left\mid \boldsymbol{Q} \right\mid \cos \alpha \implies \textcolor{red}{\alpha = \pm \frac{\pi}{2}}$