Question

Question #ad6b0

Answer 1

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

Explanation:

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`