Math Question and Answers


Question and Answers Forum

All Questions Topic List
Vector Questions
Previous in All Question Next in All Question
Previous in Vector Next in Vector
Question Number 130537 by benjo_mathlover last updated on 26/Jan/21

	Answered by EDWIN88 last updated on 26/Jan/21
	$⇔ a^→ ×(b^→ ×c^→ )=(a^→ .c^→ )b^→ −(a^→ .b^→ )c^→ ⇔ (a^→ .c^→ )b^→ −(a^→ .b^→ )c^→ =((√3)/2) b^→ +((√3)/2) c^→ we get { ((a^→ .c^→ =((√3)/2))),((a^→ .b^→ =−((√3)/2))) :} so a^→ .b^→ = ∣a^→ ∣.∣b^→ ∣ cos ϕ = −((√3)/2) ⇔ 1.1.cos ϕ=−((√3)/2) ; ϕ=((5π)/6)$
	$\Leftrightarrow \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} . \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c}$ $\Leftrightarrow (\vec{a} . \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c} = \frac{\sqrt{3}}{2} \vec{b} + \frac{\sqrt{3}}{2} \vec{c}$ $w e g e t {\begin{cases} \vec{a} . \vec{c} = \frac{\sqrt{3}}{2} \\ \vec{a} . \vec{b} = - \frac{\sqrt{3}}{2} \end{cases}$ $s o \vec{a} . \vec{b} = ∣ \vec{a} ∣ . ∣ \vec{b} ∣ \cos φ = - \frac{\sqrt{3}}{2}$ $\Leftrightarrow 1.1 . \cos φ = - \frac{\sqrt{3}}{2}; φ = \frac{5 π}{6}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum