Question

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

The answer is (8√5 i + 4√5j)/5.

Could someone please give me a detailed explanation on how to do the problem? Thanks

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,

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

Answer 2

A vector is an quantity which have both magnitude and direction.

Vector v is,     $v=\frac{8}{\sqrt{5} }i+\frac{4}{\sqrt{5} }j$

Since, in the given vector v , component in the i direction is twice the component in the j direction.

Let us consider, component in j direction is x then component in i direction will be 2x.

So,  vector    $v=2xi+xj$

Magnitude of above vector,  <v> = $\sqrt{4x^{2} +x^{2} } =4$

                                  $\sqrt{5x^{2} } =4\\\\x=\frac{4}{\sqrt{5} }$

Substituting value of x in above vector equation.

We get,    $v=\frac{8}{\sqrt{5} }i+\frac{4}{\sqrt{5} }j$

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