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Question Number 191077 by Shrinava last updated on 17/Apr/23
a^→ ∙ b^(→) = 4 b^→ ∙ c^(→) = 5 7a^(→) = 4b^(→) + 2c^(→) Find: ∣c^(→) ∣ = ?
$$\overset{\rightarrow} {\mathrm{a}}\:\centerdot\:\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{4} \\ $$$$\overset{\rightarrow} {\mathrm{b}}\:\centerdot\:\overset{\rightarrow} {\mathrm{c}}\:=\:\mathrm{5} \\ $$$$\overset{\rightarrow} {\mathrm{7a}}\:=\:\overset{\rightarrow} {\mathrm{4b}}\:+\:\overset{\rightarrow} {\mathrm{2c}}\:\: \\ $$$$\mathrm{Find}:\:\:\:\mid\overset{\rightarrow} {\mathrm{c}}\:\mid\:=\:? \\ $$
Commented by mr W last updated on 18/Apr/23
no unique solution is possible. are you sure the question is correct?
$${no}\:{unique}\:{solution}\:{is}\:{possible}. \\ $$$${are}\:{you}\:{sure}\:{the}\:{question}\:{is}\:{correct}? \\ $$
Commented by Shrinava last updated on 18/Apr/23
yes dear professor (question wrong)
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor}\:\left(\mathrm{question}\:\mathrm{wrong}\right) \\ $$
Answered by aleks041103 last updated on 18/Apr/23
⇒49a^2 =16b^2 +4c^2 +2b^→ .c^→ ⇒49a^2 =16b^2 +4c^2 +10 ⇒4c^2 =49a^2 +16b^2 −2a^→ .b^→ ⇒4c^2 =49a^2 +16b^2 −8 7a^→ .b^→ =4b^→ .b^→ +2c^→ .b^→ 28=4b^2 +10⇒b^2 =(9/2) ⇒ { ((49a^2 =82+4c^2 )),((4c^2 =49a^2 +64)) :} ⇒49a^2 =82+49a^2 +64⇒0=146 contradiction! ⇒no solution
$$\Rightarrow\mathrm{49}{a}^{\mathrm{2}} =\mathrm{16}{b}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{2}\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}} \\ $$$$\Rightarrow\mathrm{49}{a}^{\mathrm{2}} =\mathrm{16}{b}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{10} \\ $$$$\Rightarrow\mathrm{4}{c}^{\mathrm{2}} =\mathrm{49}{a}^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} −\mathrm{2}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}} \\ $$$$\Rightarrow\mathrm{4}{c}^{\mathrm{2}} =\mathrm{49}{a}^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} −\mathrm{8} \\ $$$$\mathrm{7}\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}=\mathrm{4}\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{b}}+\mathrm{2}\overset{\rightarrow} {{c}}.\overset{\rightarrow} {{b}} \\ $$$$\mathrm{28}=\mathrm{4}{b}^{\mathrm{2}} +\mathrm{10}\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{49}{a}^{\mathrm{2}} =\mathrm{82}+\mathrm{4}{c}^{\mathrm{2}} }\\{\mathrm{4}{c}^{\mathrm{2}} =\mathrm{49}{a}^{\mathrm{2}} +\mathrm{64}}\end{cases} \\ $$$$\Rightarrow\mathrm{49}{a}^{\mathrm{2}} =\mathrm{82}+\mathrm{49}{a}^{\mathrm{2}} +\mathrm{64}\Rightarrow\mathrm{0}=\mathrm{146} \\ $$$${contradiction}! \\ $$$$\Rightarrow{no}\:{solution} \\ $$
Answered by mr W last updated on 18/Apr/23
Commented by mr W last updated on 18/Apr/23
a=∣a^→ ∣ b=∣b^→ ∣ c=∣c^→ ∣ from 7a^→ =4b^→ +2c^→ : (a/(sin β))=((2c)/(7 sin α))=((4b)/(7 sin (β−α))) ⇒a=((2 sin β c)/(7 sin α)) ⇒b=((sin (β−α) c)/(2 sin α)) a^→ ∙b^→ =4 ⇒ab cos α=4 ⇒((2 sin β c)/(7 sin α))×((sin (β−α) c)/(2 sin α))×cos α=4 ⇒c^2 =((28 sin^2 α)/(cos α sin β sin (β−α))) ...(i) b^→ ∙c^→ =5 ⇒bc cos β=5 ⇒((sin (β−α) c)/(2 sin α))×c cos β=5 ⇒c^2 =((10 sin α)/(cos β sin (β−α))) ...(ii) ((28 sin^2 α)/(cos α sin β sin (β−α)))=((10 sin α)/(cos β sin (β−α))) ⇒tan α=(5/(14)) tan β put this into (ii) we get c^2 =((50)/(9 cos^2 β)) ⇒c=((5(√2))/(3 ∣cos β∣)) we see there is no unique value for c. but c_(min) =((5(√2))/3).
$${a}=\mid\overset{\rightarrow} {{a}}\mid \\ $$$${b}=\mid\overset{\rightarrow} {{b}}\mid \\ $$$${c}=\mid\overset{\rightarrow} {{c}}\mid \\ $$$${from}\:\mathrm{7}\overset{\rightarrow} {{a}}=\mathrm{4}\overset{\rightarrow} {{b}}+\mathrm{2}\overset{\rightarrow} {{c}}: \\ $$$$\frac{{a}}{\mathrm{sin}\:\beta}=\frac{\mathrm{2}{c}}{\mathrm{7}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{4}{b}}{\mathrm{7}\:\mathrm{sin}\:\left(\beta−\alpha\right)} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}\:\mathrm{sin}\:\beta\:{c}}{\mathrm{7}\:\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow{b}=\frac{\mathrm{sin}\:\left(\beta−\alpha\right)\:{c}}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$$\overset{\rightarrow} {{a}}\centerdot\overset{\rightarrow} {{b}}=\mathrm{4} \\ $$$$\Rightarrow{ab}\:\mathrm{cos}\:\alpha=\mathrm{4} \\ $$$$\Rightarrow\frac{\mathrm{2}\:\mathrm{sin}\:\beta\:{c}}{\mathrm{7}\:\mathrm{sin}\:\alpha}×\frac{\mathrm{sin}\:\left(\beta−\alpha\right)\:{c}}{\mathrm{2}\:\mathrm{sin}\:\alpha}×\mathrm{cos}\:\alpha=\mathrm{4} \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\frac{\mathrm{28}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\beta−\alpha\right)}\:\:\:…\left({i}\right) \\ $$$$\overset{\rightarrow} {{b}}\centerdot\overset{\rightarrow} {{c}}=\mathrm{5} \\ $$$$\Rightarrow{bc}\:\mathrm{cos}\:\beta=\mathrm{5} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\left(\beta−\alpha\right)\:{c}}{\mathrm{2}\:\mathrm{sin}\:\alpha}×{c}\:\mathrm{cos}\:\beta=\mathrm{5} \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\frac{\mathrm{10}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\beta\:\mathrm{sin}\:\left(\beta−\alpha\right)}\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{28}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\beta−\alpha\right)}=\frac{\mathrm{10}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\beta\:\mathrm{sin}\:\left(\beta−\alpha\right)} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{5}}{\mathrm{14}}\:\mathrm{tan}\:\beta \\ $$$${put}\:{this}\:{into}\:\left({ii}\right)\:{we}\:{get} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{50}}{\mathrm{9}\:\mathrm{cos}^{\mathrm{2}} \:\beta} \\ $$$$\Rightarrow{c}=\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{3}\:\mid\mathrm{cos}\:\beta\mid} \\ $$$${we}\:{see}\:{there}\:{is}\:{no}\:{unique}\:{value}\:{for}\:{c}. \\ $$$${but}\:{c}_{{min}} =\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{3}}. \\ $$
Answered by manolex last updated on 18/Apr/23
(.b) 7a.b=4∣b∣^2 +2b.c 28=4∣b∣^2 +10 ∣b∣^2 =(9/2) (.a) 7∣a∣^2 =4a.b+2a.c ×7 (.c) 7a.c=4b.c+2∣c∣^2 ×2 49∣a∣^2 =112+14a.c 14a.c=40+4∣c∣^2 (7∣a∣−2∣c∣)(7∣a∣+2∣c∣)=152
$$\left(.{b}\right)\:\:\:\:\mathrm{7}{a}.{b}=\mathrm{4}\mid{b}\mid^{\mathrm{2}} +\mathrm{2}{b}.{c} \\ $$$$\mathrm{28}=\mathrm{4}\mid{b}\mid^{\mathrm{2}} +\mathrm{10} \\ $$$$\mid{b}\mid^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\:\left(.{a}\right)\:\:\:\mathrm{7}\mid{a}\mid^{\mathrm{2}} =\mathrm{4}{a}.{b}+\mathrm{2}{a}.{c}\:\:×\mathrm{7} \\ $$$$\left(.{c}\right)\:\:\:\:\:\mathrm{7}{a}.{c}=\mathrm{4}{b}.{c}+\mathrm{2}\mid{c}\mid^{\mathrm{2}} \:\:\:×\mathrm{2} \\ $$$$\mathrm{49}\mid{a}\mid^{\mathrm{2}} =\mathrm{112}+\mathrm{14}{a}.{c} \\ $$$$\mathrm{14}{a}.{c}=\mathrm{40}+\mathrm{4}\mid{c}\mid^{\mathrm{2}} \\ $$$$\left(\mathrm{7}\mid{a}\mid−\mathrm{2}\mid{c}\mid\right)\left(\mathrm{7}\mid{a}\mid+\mathrm{2}\mid{c}\mid\right)=\mathrm{152} \\ $$$$ \\ $$
Commented by Shrinava last updated on 18/Apr/23
thank you so much dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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