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Question : 34 of 160
Marks:
+1,
-0
Solution:
(a+2b−c)⋅[(a−b)×(a−b−c)] =(a+2b−c)⋅(a×a−a×b−a×c −b×a+b×b+b×c) =(a+2b−c)⋅(0−a×b−a×c+a×b+0+b×c) =(a+2b−c)⋅(b×c−a×c) =a⋅(b×c)−a⋅(a×c)+2b⋅(b×c)−2b⋅(a×c) −c⋅(b×c)+c⋅(a×c) =a⋅(b×c)+2b⋅(c×a) =[]+2[bca]=[abc]+2[abc], =3[abc]].
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