a-13-b-19-a-b-24-a-b-

Question Number 169885 by depressiveshrek last updated on 11/May/22

∣ \vec{a} ∣= 13

∣ \vec{b} ∣= 19

∣ \vec{a} + \vec{b} ∣= 24

∣ \vec{a} - \vec{b} ∣= ?

Answered by mr W last updated on 11/May/22

∣ \vec{a} - \vec{b} ∣= x

x^{2} + 24^{2} = 2 \times (13^{2} + 19^{2})

\Rightarrow x = \sqrt{2 \times (13^{2} + 19^{2}) - 24^{2}} = 22 ✓

Commented by depressiveshrek last updated on 11/May/22

s o r r y w h e r e d o e s t h e 2 c o m e f r o m ?

Commented by mr W last updated on 12/May/22

Commented by mr W last updated on 11/May/22

\vec{c} = \vec{a} + \vec{b}

\vec{d} = \vec{a} - \vec{b}

a^{2} = {(\frac{d}{2})}^{2} + {(\frac{c}{2})}^{2} - 2 \times (\frac{d}{2}) \times (\frac{c}{2}) \times \cos θ

b^{2} = {(\frac{d}{2})}^{2} + {(\frac{c}{2})}^{2} + 2 \times (\frac{d}{2}) \times (\frac{c}{2}) \times \cos θ

\Rightarrow a^{2} + b^{2} = 2 \times {(\frac{d}{2})}^{2} + 2 \times {(\frac{c}{2})}^{2}

\Rightarrow c^{2} + d^{2} = 2 (a^{2} + b^{2})

Facebook Tweet Pin