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The xy-plane is the set of all points whose coordinates are of the form (x,y,0). Suppose that a vector, drawn in standard position, has it's head in the plane. Prove that this vector is perpendicular to the vector $\hat{z} = j$.

I know that two vectors can be proven to be perpendicular if their dot product is equal to zero. So, for a given vector $\overrightarrow{u}$ drawn in the xy-plane, if $y=0$, it will be perpendicular to $\hat{z}$. However, I'm confused whether this actually answered the prompt?

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    Why do you denote $\;\vec z=\vec j\;$ when its usual notation *in physics* is, as far as I know, $\;\vec z=\vec k\;$ ? I mean, the vector $\;\vec k=(0,0,1)\;$ in real three dimensional space. – 2017-02-23
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    The vector is in the $xy$ plane if $z=0.$ (not $y=0$) – 2017-02-23

1 Answers 1

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Given that you know:

  1. A general vector in the $xy$ plane is $(x,y,0)$.
  2. The vector $\hat z$ is the vector $(0,0,1)$.

What is the dot product of a general vector in the $xy$ plane and $\hat z$ equal to?