Find two unit vectors that are parallel to the xy plane

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The discussion focuses on finding two unit vectors that are parallel to the xy plane and perpendicular to the vector [1, -2, 2]. A vector parallel to the xy plane has a z component of 0, taking the form [a, b, 0]. The condition for perpendicularity to [1, -2, 2] is satisfied when the dot product equals zero, leading to the equation a - 2b = 0. The solutions include the vectors [2, 1, 0] and [4, 2, 0], with the latter being a scalar multiple of the former. However, only unit vectors are required, which limits the solutions to two specific unit vectors derived from these calculations.

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I got the answer to this question but I don't quite understand why its like that...The question is:

Find two unit vectors that are parallel to the xy plane and perpendicular to the vector [1,-2,2]
 
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Find two unit vectors that are parallel to the xy plane and perpendicular to the vector [1,-2,2].

It would be better to show us what you had done and exactly what it is that you "dont quite understand".

However, since this is very simple: Saying that a vector is "parallel to the xy plane" means that its z component i 0- you are looking for a vector of the form [a, b, 0].

Saying that the vector is "perpendicular to the vector [1, -2, 2] means that its dot product with that vector is 0: [a, b, 0].[1, -2, 2]= a- 2b= 0.

That is one equation in 2 unknowns so it have an infinite number of solutions. In particular, if you take b= 1, then a-2= 0 so a= 2.
[2, 1, 0] is a vector "parallel to the xy plane and perpendicular to the vector [1, -2, 2]". Taking b= 2, a- 4= 0 so [4, 2, 0] is another. In fact, it is just [1, -2, 0] multiplied by 2. Obviously, if one vector is "parallel to the xy plane and perpendicular to the vector [1, -2, 2]", then any multiple of it is also because it is in the same direction.
 
Plus, it says 'unit vectors'. So there's only 2 answers to this.
 

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