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Question Number 44848 by pieroo last updated on 05/Oct/18
(1)Ifa→=a1i+a2j+a3kandb→=b1i+b2j+b3kshowthati.a→×b→=|ijka1a2a3b1b2b3|ii.a→∙b→=a1b1+a2b2+a3b3(2)Ifa→=a1i+a2j+a3k,b→=b1i+b2j+b3kandc→=c1i+c2j+c3k,showthati.a→∙(b→×c→)=|a1a2a3b1b2b3c1c2c3|ii.a∙(b→×c→)=b→(a→∙c→)−c→(a→∙b→)iii.(a→×b→)×c→=b→(a→∙c→)−a→(b→∙c→)iv.a→×(b→×c→)+b→×(c→×a→)+c→×(a→×b→)=0
Commented by pieroo last updated on 05/Oct/18
pleaseIneedhelpwiththeabovequestions
Commented by pieroo last updated on 06/Oct/18
stillwaitingforsomehelp,please
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