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I have a question in which I am asked to find the equation of a plane (in three dimensions and cartesian form) when I know one point in the plane (A) and a line parallel to the plane.

At first I tried to find the normal to the plane but there are too many options (see this question).

So I looked at the first line of the solution which was:

-Obtain a vector parallel to the plane and not parallel to l, e.g. i − 2j + k

-(rest of solution)

So how do I obtain a vector parallel to the plane and not parallel to l (the line) when I know only the equation of the line and a point in the plane?

The only other information the proplem has is one other point (B), which, in a subset of the question, I used to find AB which turned out to have no point of intersection with line l. However B is not a point in the plane.

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    How is your line specified? – 2017-02-17
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    As a vector equation, r = i + j + 2k + s( 3i + j -k). – 2017-02-17
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    I think we need a lot more info about the problem itself – 2017-02-17
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    Just a parallel line and one point is not enough to uniquely determine a plane. – 2017-02-17
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    @Triatticus I have added some more info, which might have bearing, but if it's not enough I could add the whole question in. – 2017-02-20
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    The comment by mvw seems to be right, if the line is not *in* the plane then the plane is not unique. – 2017-03-03

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I know that this is over a year old now, but for anyone else who stumbles upon this post looking for help, it is currently unanswered so I'll just quickly answer it now:

You can take the position vector of the line and add it to the 'negative' of A, thus finding a vector on the plane we'll call AR. (Just vector addition). From here use vector product to cross product AR and the direction component of the line r. This will find you a vector that is perpendicular to both.

Then simply substitute what we know into the "vector equation of a plane"