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Question-130537




Question Number 130537 by benjo_mathlover last updated on 26/Jan/21
Answered by EDWIN88 last updated on 26/Jan/21
⇔ a^→ ×(b^→ ×c^→ )=(a^→ .c^→ )b^→ −(a^→ .b^→ )c^→   ⇔ (a^→ .c^→ )b^→ −(a^→ .b^→ )c^→ =((√3)/2) b^→ +((√3)/2) c^→   we get  { ((a^→ .c^→ =((√3)/2))),((a^→ .b^→ =−((√3)/2))) :}  so a^→ .b^→  = ∣a^→ ∣.∣b^→ ∣ cos ϕ = −((√3)/2)  ⇔ 1.1.cos ϕ=−((√3)/2) ; ϕ=((5π)/6)
$$\Leftrightarrow\:\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{c}}\right)=\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}\right)\overset{\rightarrow} {{c}} \\ $$$$\Leftrightarrow\:\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{b}}−\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}\right)\overset{\rightarrow} {{c}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\overset{\rightarrow} {{b}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\overset{\rightarrow} {{c}} \\ $$$${we}\:{get}\:\begin{cases}{\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$${so}\:\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}\:=\:\mid\overset{\rightarrow} {{a}}\mid.\mid\overset{\rightarrow} {{b}}\mid\:\mathrm{cos}\:\varphi\:=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{1}.\mathrm{1}.\mathrm{cos}\:\varphi=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:;\:\varphi=\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

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