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Let-vector-a-b-and-c-such-that-a-b-c-2-and-a-a-c-b-0-find-the-acute-angle-between-a-and-c-

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Question Number 136300 by liberty last updated on 20/Mar/21
Let vector a^→ , b^→ and c^→ such that ∣a^→ ∣=∣b^→ ∣=((∣c^→ ∣)/2) and a^→ ×(a^→ ×c^→ )+b^→ =0 find the acute angle between a^→ and c^→ .
$${Let}\:{vector}\:\overset{\rightarrow} {{a}}\:,\:\overset{\rightarrow} {{b}}\:{and}\:\overset{\rightarrow} {{c}}\:{such}\:{that} \\ $$$$\mid\overset{\rightarrow} {{a}}\mid=\mid\overset{\rightarrow} {{b}}\mid=\frac{\mid\overset{\rightarrow} {{c}}\mid}{\mathrm{2}}\:{and}\:\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{c}}\right)+\overset{\rightarrow} {{b}}=\mathrm{0} \\ $$$${find}\:{the}\:{acute}\:{angle}\:{between}\:\overset{\rightarrow} {{a}}\:{and}\:\overset{\rightarrow} {{c}}\:. \\ $$
Answered by EDWIN88 last updated on 20/Mar/21
let ∣a^→ ∣=∣b^→ ∣ = k then ∣c^→ ∣ = 2k and let the angle between a^→ & c^→ is β ⇒ a^→ ×(a^→ ×c^→ )=(a^→ .c^→ )a^→ −(a^→ .a^→ )c^→ ⇒a^→ ×(a^→ ×c^→ ) = (2k^2 cos β)a^→ −k^2 c^→ Then ⇒b^→ = k^2 (c^→ −2cos β a^→ ) let cos β = m ⇒ ∣b^→ ∣ = k^2 ∣c^→ −2m a^→ ∣ ⇒k = k^2 (√(4k^2 +4m^2 .k^2 −8m^2 k^2 )) ⇒(1/k^2 ) = (√(4−4m^2 )) ; 4m^2 = 4 −(1/k^4 ) ⇒ cos β = (√((4k^4 −1)/(4k^4 ))) ; β = arccos (((√(4k^4 −1))/(2k))) If k=1 ⇒β = arccos (((√3)/2)) → { ((β=30°)),((β=150°)) :}
$$\mathrm{let}\:\mid\overset{\rightarrow} {{a}}\mid=\mid\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{k}\:\mathrm{then}\:\mid\overset{\rightarrow} {{c}}\mid\:\:=\:\mathrm{2k}\:\mathrm{and}\:\mathrm{let}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {{a}}\:\&\:\overset{\rightarrow} {{c}}\:\mathrm{is}\:\beta \\ $$$$\Rightarrow\:\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{c}}\right)=\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{c}}\right)\overset{\rightarrow} {{a}}−\left(\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{a}}\right)\overset{\rightarrow} {{c}} \\ $$$$\Rightarrow\overset{\rightarrow} {{a}}×\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{c}}\right)\:=\:\left(\mathrm{2k}^{\mathrm{2}} \:\mathrm{cos}\:\beta\right)\overset{\rightarrow} {{a}}−\mathrm{k}^{\mathrm{2}} \:\overset{\rightarrow} {{c}} \\ $$$$\mathrm{Then}\:\Rightarrow\overset{\rightarrow} {{b}}\:=\:\mathrm{k}^{\mathrm{2}} \:\left(\overset{\rightarrow} {{c}}−\mathrm{2cos}\:\beta\:\overset{\rightarrow} {{a}}\:\right) \\ $$$$\mathrm{let}\:\mathrm{cos}\:\beta\:=\:\mathrm{m} \\ $$$$\Rightarrow\:\mid\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{k}^{\mathrm{2}} \:\mid\overset{\rightarrow} {{c}}−\mathrm{2m}\:\overset{\rightarrow} {{a}}\mid\: \\ $$$$\Rightarrow\mathrm{k}\:=\:\mathrm{k}^{\mathrm{2}} \:\sqrt{\mathrm{4k}^{\mathrm{2}} +\mathrm{4m}^{\mathrm{2}} .\mathrm{k}^{\mathrm{2}} −\mathrm{8m}^{\mathrm{2}} \mathrm{k}^{\mathrm{2}} }\: \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{4}−\mathrm{4m}^{\mathrm{2}} }\:;\:\mathrm{4m}^{\mathrm{2}} \:=\:\mathrm{4}\:−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{4}} } \\ $$$$\Rightarrow\:\mathrm{cos}\:\beta\:=\:\sqrt{\frac{\mathrm{4k}^{\mathrm{4}} −\mathrm{1}}{\mathrm{4k}^{\mathrm{4}} }}\:;\:\beta\:=\:\mathrm{arccos}\:\left(\frac{\sqrt{\mathrm{4k}^{\mathrm{4}} −\mathrm{1}}}{\mathrm{2k}}\right) \\ $$$$\mathrm{If}\:\mathrm{k}=\mathrm{1}\:\Rightarrow\beta\:=\:\mathrm{arccos}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:\rightarrow\begin{cases}{\beta=\mathrm{30}°}\\{\beta=\mathrm{150}°}\end{cases} \\ $$$$ \\ $$
Answered by mr W last updated on 20/Mar/21
Commented by mr W last updated on 20/Mar/21
let ∣a∣=∣b∣=s, then ∣c∣=2s ∣a×c∣=∣a∣∣c∣ sin θ ∣a×(a×c)∣=∣a∣∣a∣∣c∣ sin θ×sin (π/2) =2s^3 sin θ since a×(a×c)+b=0, ⇒∣a×(a×c)∣=∣b∣ ⇒2s^3 sin θ=s ⇒sin θ=(1/(2s^2 )) ⇒θ=sin^(−1) (1/(2s^2 )) or π−sin^(−1) (1/(2s^2 )) i.e. the acute angle between a and c is sin^(−1) (1/(2s^2 )) if s=1, then θ=(π/6)
$${let}\:\mid\boldsymbol{{a}}\mid=\mid\boldsymbol{{b}}\mid={s},\:{then}\:\mid\boldsymbol{{c}}\mid=\mathrm{2}{s} \\ $$$$\mid\boldsymbol{{a}}×\boldsymbol{{c}}\mid=\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{c}}\mid\:\mathrm{sin}\:\theta \\ $$$$\mid\boldsymbol{{a}}×\left(\boldsymbol{{a}}×\boldsymbol{{c}}\right)\mid=\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{a}}\mid\mid\boldsymbol{{c}}\mid\:\mathrm{sin}\:\theta×\mathrm{sin}\:\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{s}^{\mathrm{3}} \:\mathrm{sin}\:\theta \\ $$$${since}\:\boldsymbol{{a}}×\left(\boldsymbol{{a}}×\boldsymbol{{c}}\right)+\boldsymbol{{b}}=\mathrm{0}, \\ $$$$\Rightarrow\mid\boldsymbol{{a}}×\left(\boldsymbol{{a}}×\boldsymbol{{c}}\right)\mid=\mid\boldsymbol{{b}}\mid \\ $$$$\Rightarrow\mathrm{2}{s}^{\mathrm{3}} \:\mathrm{sin}\:\theta={s} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{s}^{\mathrm{2}} } \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{s}^{\mathrm{2}} }\:{or}\:\pi−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{s}^{\mathrm{2}} } \\ $$$${i}.{e}.\:{the}\:{acute}\:{angle}\:{between}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{c}}\:{is} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{s}^{\mathrm{2}} } \\ $$$${if}\:{s}=\mathrm{1},\:{then}\:\theta=\frac{\pi}{\mathrm{6}} \\ $$
Answered by mnjuly1970 last updated on 20/Mar/21
a⊥a×c : a×(a×c)=−b ⇒∣a×(a×c)∣=∣b∣ ∣a∣∣a×c∣=∣b∣ ∣a∣∣a∣∣c∣sin(ϕ)=∣b∣ sin(ϕ)=((∣b∣)/(2∣b∣^2 ∣b∣))=(1/(2∣b∣^2 )) ∣b∣#0 ,0<(1/(∣b∣^2 ))≤2⇒∣b∣≥(1/( (√2))) ϕ=sin^(−1) ((1/(2∣b∣^2 ))) ϕ=π−sin^(−1) ((1/(2∣b∣^2 )))
$$\:\:{a}\bot{a}×{c}\:: \\ $$$$\:\:\:{a}×\left({a}×{c}\right)=−{b} \\ $$$$\:\:\Rightarrow\mid{a}×\left({a}×{c}\right)\mid=\mid{b}\mid \\ $$$$\:\:\:\:\:\mid{a}\mid\mid{a}×{c}\mid=\mid{b}\mid \\ $$$$\:\:\:\:\:\:\mid{a}\mid\mid{a}\mid\mid{c}\mid{sin}\left(\varphi\right)=\mid{b}\mid \\ $$$$\:\:\:\:\:\:\:\:{sin}\left(\varphi\right)=\frac{\mid{b}\mid}{\mathrm{2}\mid{b}\mid^{\mathrm{2}} \mid{b}\mid}=\frac{\mathrm{1}}{\mathrm{2}\mid{b}\mid^{\mathrm{2}} } \\ $$$$\:\:\:\mid{b}\mid#\mathrm{0}\:,\mathrm{0}<\frac{\mathrm{1}}{\mid{b}\mid^{\mathrm{2}} }\leqslant\mathrm{2}\Rightarrow\mid{b}\mid\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\varphi={sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}\mid{b}\mid^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\varphi=\pi−{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}\mid{b}\mid^{\mathrm{2}} }\right) \\ $$$$ \\ $$

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