a^→ ∙ b^(→) = 4 b^→ ∙ c^(→) = 5 7a^(→) = 4b^(→) + 2c^(→) Find: ∣c^(→) ∣ = ?


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Question Number 191077 by Shrinava last updated on 17/Apr/23

$\vec{a} \cdot \vec{b} = 4$ $\vec{b} \cdot \vec{c} = 5$ $\vec{7 a} = \vec{4 b} + \vec{2 c}$ $Find : ∣ \vec{c} ∣ = ?$
Commented by mr W last updated on 18/Apr/23

$n o u n i q u e s o l u t i o n i s p o s s i b l e .$ $a r e y o u s u r e t h e q u e s t i o n i s c o r r e c t ?$
Commented by Shrinava last updated on 18/Apr/23

$yes dear professor (question wrong)$
	Answered by aleks041103 last updated on 18/Apr/23
	$⇒49a^2 =16b^2 +4c^2 +2b^→ .c^→ ⇒49a^2 =16b^2 +4c^2 +10 ⇒4c^2 =49a^2 +16b^2 −2a^→ .b^→ ⇒4c^2 =49a^2 +16b^2 −8 7a^→ .b^→ =4b^→ .b^→ +2c^→ .b^→ 28=4b^2 +10⇒b^2 =(9/2) ⇒ { ((49a^2 =82+4c^2 )),((4c^2 =49a^2 +64)) :} ⇒49a^2 =82+49a^2 +64⇒0=146 contradiction! ⇒no solution$
	$\Rightarrow 49 a^{2} = 16 b^{2} + 4 c^{2} + 2 \vec{b} . \vec{c}$ $\Rightarrow 49 a^{2} = 16 b^{2} + 4 c^{2} + 10$ $\Rightarrow 4 c^{2} = 49 a^{2} + 16 b^{2} - 2 \vec{a} . \vec{b}$ $\Rightarrow 4 c^{2} = 49 a^{2} + 16 b^{2} - 8$ $7 \vec{a} . \vec{b} = 4 \vec{b} . \vec{b} + 2 \vec{c} . \vec{b}$ $28 = 4 b^{2} + 10 \Rightarrow b^{2} = \frac{9}{2}$ $\Rightarrow {\begin{cases} 49 a^{2} = 82 + 4 c^{2} \\ 4 c^{2} = 49 a^{2} + 64 \end{cases}$ $\Rightarrow 49 a^{2} = 82 + 49 a^{2} + 64 \Rightarrow 0 = 146$ $c o n t r a d i c t i o n!$ $\Rightarrow n o s o l u t i o n$
	Answered by mr W last updated on 18/Apr/23

	Commented by mr W last updated on 18/Apr/23

	$a =∣ \vec{a} ∣$ $b =∣ \vec{b} ∣$ $c =∣ \vec{c} ∣$ $f r o m 7 \vec{a} = 4 \vec{b} + 2 \vec{c} :$ $\frac{a}{\sin β} = \frac{2 c}{7 \sin α} = \frac{4 b}{7 \sin (β - α)}$ $\Rightarrow a = \frac{2 \sin β c}{7 \sin α}$ $\Rightarrow b = \frac{\sin (β - α) c}{2 \sin α}$ $\vec{a} \cdot \vec{b} = 4$ $\Rightarrow a b \cos α = 4$ $\Rightarrow \frac{2 \sin β c}{7 \sin α} \times \frac{\sin (β - α) c}{2 \sin α} \times \cos α = 4$ $\Rightarrow c^{2} = \frac{28 \sin^{2} α}{\cos α \sin β \sin (β - α)} . . . (i)$ $\vec{b} \cdot \vec{c} = 5$ $\Rightarrow b c \cos β = 5$ $\Rightarrow \frac{\sin (β - α) c}{2 \sin α} \times c \cos β = 5$ $\Rightarrow c^{2} = \frac{10 \sin α}{\cos β \sin (β - α)} . . . (i i)$ $\frac{28 \sin^{2} α}{\cos α \sin β \sin (β - α)} = \frac{10 \sin α}{\cos β \sin (β - α)}$ $\Rightarrow \tan α = \frac{5}{14} \tan β$ $p u t t h i s i n t o (i i) w e g e t$ $c^{2} = \frac{50}{9 \cos^{2} β}$ $\Rightarrow c = \frac{5 \sqrt{2}}{3 ∣ \cos β ∣}$ $w e s e e t h e r e i s n o u n i q u e v a l u e f o r c .$ $b u t c_{m i n} = \frac{5 \sqrt{2}}{3} .$
	Answered by manolex last updated on 18/Apr/23

	$(. b) 7 a . b = 4 ∣ b ∣^{2} + 2 b . c$ $28 = 4 ∣ b ∣^{2} + 10$ $∣ b ∣^{2} = \frac{9}{2}$ $(. a) 7 ∣ a ∣^{2} = 4 a . b + 2 a . c \times 7$ $(. c) 7 a . c = 4 b . c + 2 ∣ c ∣^{2} \times 2$ $49 ∣ a ∣^{2} = 112 + 14 a . c$ $14 a . c = 40 + 4 ∣ c ∣^{2}$ $(7 ∣ a ∣ - 2 ∣ c ∣) (7 ∣ a ∣ + 2 ∣ c ∣) = 152$
	Commented by Shrinava last updated on 18/Apr/23

	$thank you so much dear professors$
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