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Question Number 119572 by zakirullah last updated on 25/Oct/20
Commented by bemath last updated on 25/Oct/20
(i)method1a→∥b→ifa→×b→=0→⇒a→×b→=|−12−32−46|=(12−12,−6−(−6),4−4)=(0,0,0)
(i)method2a→∥b→ifa1b1=a2b2=a3b3then−12=2−4=−36
Commented by zakirullah last updated on 25/Oct/20
youarethegreat!
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