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

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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:

$\frac{| | a | | \cdot b}{| | b | |}$ $(||a||*b)/||b||$

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

$| | v | | = 9$ $||v||=9$

$\frac{9 (2 + 5)}{\sqrt{29}}$ $(9(2+5))/sqrt(29)$

$\frac{18}{\sqrt{29}} ˆ i + \frac{45}{\sqrt{29}} ˆ j$ $18/sqrt(29)hati+45/sqrt(29)hatj$ 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>u=<2,5>?