Vector A=i- 2j+ k, what will be the unit vector parallel to XY plane and perpendicular to vector A?

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Answer 1 · 2018-06-05T08:37:42

#1/sqrt5 ( 2,1,0 )#

The x-y plane has unit normal vector: #bb hat k = (0,0,1)#

The vector you want (call it #bb V#) is perp to both #bb hat k# and #bb A = (1, -2, 1)#

That is: #bb V = bb A times bb hat k#

#= det [(bb hat i , bb hat j, bb hat k),(1, -2 ,1),(0,0,1)]#

#= bb hat i (-2) - bb hat j (1) + bb hat k (0) = (-2, -1, 0)#

We can equally use #bb V = (2,1,0)# to avoid the minus signs.

For the unit vector:

#bb hat V = (bb V)/(abs bb V) = ( (( 2,1,0 )) )/sqrt(2^2 + 1^2 + 0^2) =1/sqrt5 ( 2,1,0 )#