a-b-4-b-c-5-7a-4b-2c-Find-c-

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

\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)} \dots (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)} \dots (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