How do you find the vector v with the given magnitude of 9 and in the same direction as $u=<2,5>$ ?

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Answer 1 · 2017-12-17T19:34:46

See below.

A vector a with given magnitude in the direction of b, is given by:

$(||a||*b)/||b||$

$||u||=sqrt((2)^2+(5)^2)=sqrt(29)$

$||v||=9$

$(9(2+5))/sqrt(29)$

$18/sqrt(29)hati+45/sqrt(29)hatj$ or $((18/sqrt(29)),(45/sqrt(29)))$

How do you find the vector v with the given magnitude of 9 and in the same direction as u=<2,5>u=<2,5>?