Question #ad6b0

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Answer 1 · 2017-06-03T06:36:12

The vector is $veca=2sqrt5veci+sqrt5vecj-5veckj$

Let the $veca$ be

$veca=a_xveci+a_yvecj+a_zveck$

We are given

$a_x=2a_y$

and

$cos135º =a_z/(||veca||)$

$||veca||=sqrt(a_x^2+a_y^2+a_z^2)=5sqrt2$

$cos 135=-sqrt2/2=a_z/(5sqrt2)$

$=>$ , $a_z=-sqrt2/2*5sqrt2=-5$

Therefore,

$a_x^2+a_y^2+a_z^2=(5sqrt2)^2$

$a_x^2+a_y^2=50-25=25$

$4a_y^2+a_y^2=25$

$a_y^2=25/5=5$

$a_y=sqrt5$

$a_x=2sqrt5$

So,

$veca=2sqrt5veci+sqrt5vecj-5veckj$