Question #807f7

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Answer 1 · 2017-03-17T19:05:12

#mathbf u_1 = ((4/5),(3/5))#

#mathbf u_2 = ((-4/5),(-3/5))#

I think what you're saying is that #(4,3)# is the normal vector to the plane you mention.

If that is so, the first unit vector is:

#mathbf u_1 = 1/sqrt(4^2 + 3^2)((4),(3))#

#= ((4/5),(3/5))#

The other unit vector will point in the opposite direction, ie:

#mathbf u_2 = ((-4/5),(-3/5))#