(a^→ ×b^→ )×(a^→ )=? how is the solution


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Question Number 207332 by mustafazaheen last updated on 12/May/24

$(\vec{a} \times \vec{b}) \times (\vec{a}) = ?$ $how is the solution$
Commented by Ghisom last updated on 12/May/24

$((\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix}) \times (\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix})) \times (\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix}) =$ $= (\begin{matrix} a_{2} b_{3} - a_{3} b_{2} \\ a_{3} b_{1} - a_{1} b_{3} \\ a_{1} b_{2} - a_{2} b_{1} \end{matrix}) \times (\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix}) =$ $= (\begin{matrix} (a_{2}^{2} + a_{3}^{2}) b_{1} - a_{1} (a_{2} b_{2} + a_{3} b_{3}) \\ (a_{1}^{2} + a_{3}^{2}) b_{2} - a_{2} (a_{1} b_{1} + a_{3} b_{3}) \\ (a_{1}^{2} + a_{2}^{2}) b_{3} - a_{3} (a_{1} b_{1} + a_{2} b_{2}) \end{matrix})$
	Answered by aleks041103 last updated on 12/May/24

	$(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} . \vec{c}) \vec{b} - (\vec{b} . \vec{c}) \vec{a}$ $\Rightarrow (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} . \vec{a}) \vec{b} - (\vec{b} . \vec{a}) \vec{a} =$ $=∣ \vec{a} ∣^{2} \vec{b} - (\vec{a} . \vec{b}) \vec{a}$
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