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

$\frac{1}{\sqrt{5}} (2, 1, 0)$ $1/sqrt5 ( 2,1,0 )$

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

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

That is: $V = A \times ˆ k$ $bb V = bb A times bb hat k$

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

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

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

For the unit vector:

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