∣a^→ ∣=13 ∣b^→ ∣=19 ∣a^→ +b^→ ∣=24 ∣a^→ −b^→ ∣=?


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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

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