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Question : 36 of 160
Marks:
+1,
-0
Solution:
(a×b)×c=(a⋅c)b−(b⋅c)a =[(2−3+)⋅(−)(+2−3‌)] −[(+2−3)⋅(−)](2−3+) =5(+2−3)+(2−3+) =7+7−14 (a×b)×c⟂d‌ ⇒‌‌[(a×b)×c]⋅d=0 ‌(7+7−14)⋅(++x)=0 ⇒‌‌7+7−14x=0‌‌‌ or ‌‌‌x=1
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