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A vector parallel to 5i^+2j^ and having a magnitude of 52 is

Divya , 7 Years ago
Grade 10
anser 1 Answers
Vikas TU

Last Activity: 7 Years ago

Dear Student,
Let, A=5i+2j
|A|=52
=>√(5^2+2^2+2X5X2cos) = 52
=>cos=133.75 which is not possible as cos value can range from -1 to 1. Hence the value of magnitude should be something different.
Let, A=5i+2j
|A|=52
=>√(5^2+2^2+2X5X2cos) = 52
=>cos=133.75 which is not possible as cos value can range from -1 to 1. Hence the value of magnitude should be something different.
Here  is the thetha or angle.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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