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I'm beginning studying 2-variables function on my own.

What is the equation of a plane passing through the line of equation

$$x +y-1=0$$

and perpendicular to the xy plane?

Can you please give me the solution and a quick reference?

Thanks

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    There are many ways to find and represent the equation of a plane. What have you tried, so that someone can help you in a way that you can follow. – 2017-01-10
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    @Paul Nothing in particular – 2017-01-10
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    @f126ck There is no unique solution to your problem. Are you sure there are other restrictions you didn't mention? – 2017-01-10
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    @Paul I want the equation of a plane whose intersection with the xy plane is a line of equation $$x+y-1=0$$ and such plane has to be perpendicular to the xy plane...I can imagine only a plane as solution..Am I wrong?I'm a beginner... – 2017-01-10
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    I did that and you did not recognize it. I don't know what you know already, which is the problem. By showing what you have tried I can see where to pitch an answer, but without that it is difficult to help. What do you understand by "the equation of a plane" and how do you normally find it? – 2017-01-10
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    you don't need to actually calculate anything to find the plane, just recognize what you are looking at. – 2017-01-10

1 Answers 1

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First $x+y_1=0$, in $\mathbf R^3$ is not an equation of a line, but an equation of a plane. I suppose you mean the line with equations $$\begin{cases}x+y-1=0,\\z=0.\end{cases}$$ The answer is obviously … $\;x+y-1=0$, since the $xy$-plane has equation $z=0$ so that these planes have normal vectors $n_1=(1,1,0)$ and $n_2=(0,0,1)$ repectively, and these normal vectors are orthogonal.