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Question Number 207332 by mustafazaheen last updated on 12/May/24
(a^→ ×b^→ )×(a^→ )=?  how is the solution
(a×b)×(a)=?howisthesolution
Commented by Ghisom last updated on 12/May/24
( ((a_1 ),(a_2 ),(a_3 ) ) × ((b_1 ),(b_2 ),(b_3 ) ) )× ((a_1 ),(a_2 ),(a_3 ) ) =  = (((a_2 b_3 −a_3 b_2 )),((a_3 b_1 −a_1 b_3 )),((a_1 b_2 −a_2 b_1 )) ) × ((a_1 ),(a_2 ),(a_3 ) ) =  = ((((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 ))) )
((a1a2a3)×(b1b2b3))×(a1a2a3)==(a2b3a3b2a3b1a1b3a1b2a2b1)×(a1a2a3)==((a22+a32)b1a1(a2b2+a3b3)(a12+a32)b2a2(a1b1+a3b3)(a12+a22)b3a3(a1b1+a2b2))
Answered by aleks041103 last updated on 12/May/24
(a^(→) ×b^(→) )×c^(→) =(a^(→) .c^(→) )b^(→) −(b^(→) .c^(→) )a^(→)   ⇒(a^(→) ×b^(→) )×a^(→) =(a^(→) .a^(→) )b^(→) −(b^(→) .a^(→) )a^(→) =  =∣a^(→) ∣^2 b^(→) −(a^(→) .b^(→) )a^(→)
(a×b)×c=(a.c)b(b.c)a(a×b)×a=(a.a)b(b.a)a==∣a2b(a.b)a

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