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The maximum and minimum magnitude of 2i+3j and 5i-2j
The maximum and minimum magnitude of 2i+3j and 5i-2j


3 years ago

Shailendra Kumar Sharma
188 Points
							Two vectors A and B can give the resultant as R2=A2+B2+2ABcos(theta)So the maximum is R=A+B and minimum is A-BNow 2i+3j and 5i-2j has magnitude of $\sqrt{4+9}$  =$\sqrt{13}}$and $\sqrt{25+4}$  =$\sqrt{29}$respectively  so maximum will be  A+B=  $\sqrt{13}}$+$\sqrt{29}$=8.99 and minimum be 1.78

3 years ago
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