help with vectors?

Question

the question is:
find a vector whose magnitude is 4 and whose component in the i direction is twice the component in the j direction..
i keep on getting 4/sqrt(3) but the real answer is 4sqrt(5)/5 and i dont get how you get that

Answer 1

(i and j are the components in those directions)

i = 2j

and, by pythagorean

i^2 + j^2 = d^2

d, the distance or magnitude, is 4.

so

i^2 + j^2 = 16

plug in i = 2j

4j^2 + j^2 = 16

5j^2 = 16

j = 4/sqrt(5)

= 4sqrt(5)/5

Answer 2

We want a vector of the form v = ai + 2aj such that:

a² + (2a)² = 4²
a² + 4a² = 16
5a² = 16
a² = 16/5

a = 4/√5
2a = 8/√5

The vector is:

v = (4/√5)i + (8/√5)j

Answer 3

I think I remember how to do this....

sqrt(i^2 + j^2) = 4
and
i = 2j

sqrt((2j)^2 + j^2) = 4
4j^2 + j^2 = 16
5j^2 = 16
j^2 = 16/5
j = sqrt(16/5) = 1.7889

i = 2j = 2*sqrt(16/5) = 3.5778