Question #ad6b0

1 Answer
Jun 3, 2017

Answer:

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#