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

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

$$\mid\overset{\rightarrow} {{a}}\mid=\mathrm{13} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid=\mathrm{19} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\mathrm{24} \\ $$$$\mid\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\mid=? \\ $$

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

∣a^→ −b^→ ∣=x x^2 +24^2 =2×(13^2 +19^2 ) ⇒x=(√(2×(13^2 +19^2 )−24^2 ))=22 ✓

$$\mid\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\mid={x} \\ $$$${x}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} =\mathrm{2}×\left(\mathrm{13}^{\mathrm{2}} +\mathrm{19}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}×\left(\mathrm{13}^{\mathrm{2}} +\mathrm{19}^{\mathrm{2}} \right)−\mathrm{24}^{\mathrm{2}} }=\mathrm{22}\:\checkmark \\ $$

Commented by depressiveshrek last updated on 11/May/22

sorry where does the 2 come from?

$${sorry}\:{where}\:{does}\:{the}\:\mathrm{2}\:{come}\:{from}? \\ $$

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

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

c^→ =a^→ +b^→ d^→ =a^→ −b^→ a^2 =((d/2))^2 +((c/2))^2 −2×((d/2))×((c/2))×cos θ b^2 =((d/2))^2 +((c/2))^2 +2×((d/2))×((c/2))×cos θ ⇒a^2 +b^2 =2×((d/2))^2 +2×((c/2))^2 ⇒c^2 +d^2 =2(a^2 +b^2 )

$$\overset{\rightarrow} {{c}}=\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}} \\ $$$$\overset{\rightarrow} {{d}}=\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}} \\ $$$${a}^{\mathrm{2}} =\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}×\left(\frac{{d}}{\mathrm{2}}\right)×\left(\frac{{c}}{\mathrm{2}}\right)×\mathrm{cos}\:\theta \\ $$$${b}^{\mathrm{2}} =\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}×\left(\frac{{d}}{\mathrm{2}}\right)×\left(\frac{{c}}{\mathrm{2}}\right)×\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}×\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}×\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$

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