Question

Find a vector v whose magnitude is 4 and whose component in the i direction is twice the component in the j direction.

Answer 1

Start with

$\vec v=x\,\vec\imath+y\,\vec\jmath$

as a template for the vector $\vec v$. Its magnitude is 4, so

$\|\vec v\|=\sqrt{x^2+y^2}=4$

Its component in the $\vec\imath$ direction is twice the component in the $\vec\jmath$ direction, which means

$x=2y$

So we have

$\sqrt{(2y)^2+y^2}=\sqrt{5y^2}=4\implies y^2=\dfrac{16}5\implies y=\pm\dfrac4{\sqrt5}$

and

$x=\pm\dfrac8{\sqrt5}$

Lastly, rationalize the denominator:

$\dfrac1{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{\sqrt5}5$

So we end up with two possible answers,

$\vec v=\pm\left(\dfrac{8\sqrt5}5\,\vec\imath+\dfrac{4\sqrt5}5\,\vec\jmath\right)$