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-

Question Number 136300 by liberty last updated on 20/Mar/21

L e t v e c t o r \vec{a}, \vec{b} a n d \vec{c} s u c h t h a t

∣ \vec{a} ∣=∣ \vec{b} ∣= \frac{∣ \vec{c} ∣}{2} a n d \vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = 0

f i n d t h e a c u t e a n g l e b e t w e e n \vec{a} a n d \vec{c} .

Answered by EDWIN88 last updated on 20/Mar/21

let ∣ \vec{a} ∣=∣ \vec{b} ∣ = k then ∣ \vec{c} ∣ = 2 k and let the

angle between \vec{a} & \vec{c} is β

\Rightarrow \vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} . \vec{c}) \vec{a} - (\vec{a} . \vec{a}) \vec{c}

\Rightarrow \vec{a} \times (\vec{a} \times \vec{c}) = ({2 k}^{2} \cos β) \vec{a} - k^{2} \vec{c}

Then \Rightarrow \vec{b} = k^{2} (\vec{c} - 2 \cos β \vec{a})

let \cos β = m

\Rightarrow ∣ \vec{b} ∣ = k^{2} ∣ \vec{c} - 2 m \vec{a} ∣

\Rightarrow k = k^{2} \sqrt{{4 k}^{2} + {4 m}^{2} . k^{2} - {8 m}^{2} k^{2}}

\Rightarrow \frac{1}{k^{2}} = \sqrt{4 - {4 m}^{2}}; {4 m}^{2} = 4 - \frac{1}{k^{4}}

\Rightarrow \cos β = \sqrt{\frac{{4 k}^{4} - 1}{{4 k}^{4}}}; β = \arccos (\frac{\sqrt{{4 k}^{4} - 1}}{2 k})

If k = 1 \Rightarrow β = \arccos (\frac{\sqrt{3}}{2}) \to {\begin{cases} β = 30 ° \\ β = 150 ° \end{cases}

Answered by mr W last updated on 20/Mar/21

Commented by mr W last updated on 20/Mar/21

l e t ∣ a ∣=∣ b ∣= s, t h e n ∣ c ∣= 2 s

∣ a \times c ∣=∣ a ∣∣ c ∣ \sin θ

∣ a \times (a \times c) ∣=∣ a ∣∣ a ∣∣ c ∣ \sin θ \times \sin \frac{π}{2}

= 2 s^{3} \sin θ

s i n c e a \times (a \times c) + b = 0,

\Rightarrow∣ a \times (a \times c) ∣=∣ b ∣

\Rightarrow 2 s^{3} \sin θ = s

\Rightarrow \sin θ = \frac{1}{2 s^{2}}

\Rightarrow θ = \sin^{- 1} \frac{1}{2 s^{2}} o r π - \sin^{- 1} \frac{1}{2 s^{2}}

i . e . t h e a c u t e a n g l e b e t w e e n a a n d c i s

\sin^{- 1} \frac{1}{2 s^{2}}

i f s = 1, t h e n θ = \frac{π}{6}

Answered by mnjuly1970 last updated on 20/Mar/21

a ⊥ a \times c :

a \times (a \times c) = - b

\Rightarrow∣ a \times (a \times c) ∣=∣ b ∣

∣ a ∣∣ a \times c ∣=∣ b ∣

∣ a ∣∣ a ∣∣ c ∣ s i n (φ) =∣ b ∣

s i n (φ) = \frac{∣ b ∣}{2 ∣ b ∣^{2} ∣ b ∣} = \frac{1}{2 ∣ b ∣^{2}}

You can't use 'macro parameter character #' in math mode

φ = {s i n}^{- 1} (\frac{1}{2 ∣ b ∣^{2}})

φ = π - {s i n}^{- 1} (\frac{1}{2 ∣ b ∣^{2}})