How do you find the vector v with the given magnitude of 5 and in the same direction as #u=<3,3>#?

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Answer 1 · 2017-05-03T18:30:27

Please see the explanation.

Given: #|vecv|= 5#

We know the formula for the magnitude

#|vecv|= sqrt(x^2+y^2)" [1]"#

This problem is made simple by the fact that the x component of the vector #vecu# is the same as the y component.

#y = x#

Substitute x for y into equation [1]:

#|vecv|= sqrt(x^2+x^2)" [2]"#

Substitute 5 for the magnitude into equation [2]:

#5= sqrt(x^2+x^2)#

#25= 2x^2#

#25/2=x^2#

#x = 5sqrt2/2#

We know that #y = x#:

#y = 5sqrt(2)/2#

The vector is #< 5sqrt(2)/2, 5sqrt(2)/2 >#