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I understand that since its parallel to the xy plane then the z value would be zero but not sure about how to get the equation itself

Maro
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    You have a point. All that is left to find is the direction vector for the line. Since the line should be parallel to the xy-plane, any vector lying in the xy-plane will work fine as your direction vector. For example $\mathbf{v}=\mathbf{i}$. The equation of a line in space can be represented by the following vector-valued function: $\mathbf{r}(t)=(x_0,y_0,x_0)+t\mathbf{v}$. So, the equation of one possible line that satisfies the conditions laid out in the problem statement would be: $$\mathbf{r}(t)=(0,0,1)+t\mathbf{i}=t\mathbf{i}+\mathbf{k}.$$ – Michael Rybkin Jun 09 '19 at 06:21
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    Please avoid 'I have no clue' questions. Have you tried writing down the equation of a line in vector form, and then applying that to the question. – Toby Mak Jun 09 '19 at 07:09
  • Essentially a duplicate of https://math.stackexchange.com/q/477885/265466. – amd Jun 09 '19 at 08:17

2 Answers2

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There are infinitely many such lines. If we choose any direction parallel to the $xy$-plane, using a vector in this plane (such as $(\cos \theta, \sin \theta, 0)$), we simply obtain the parametric equation $(0,0,1) + \lambda (\cos \theta, \sin \theta, 0)$ for $\lambda \in \mathbb{R}$.

auscrypt
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The directions parallel to $xy$-plane are all such that $z=0$ in the direction-determining vector of the line.

So you'd have $\begin{pmatrix}0\\0\\1\end{pmatrix}+t\begin{pmatrix}a\\b\\0\end{pmatrix}$ for $a,b\in\mathbb{R}$

Saketh Malyala
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