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If you have a vector line m with equation $r = i + j + 2k + s( 3i + j - k)$ how do you find the normal?

If you presume that normal is $n = ( n i + o j + p k )$ then

$$3n + o -p = 0 $$

but this leaves you with too many options $\left \lbrace (0,0,0), (1,0,3), (0,1,1), ....\right \rbrace$ is it possible to find the normal?

(This is asked in relation to when I am trying to find the equation of a plane and have a line m and a point A, if I can find the normal to line $m$ I can use $n \cdot a=n \cdot r.$)

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    "the" normal is not well defined ! There is a normal vector plane and every nonzero vector in it can be considered as "a" normal vector to the line. – 2017-02-15
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    What exactly is the original question? – 2017-02-17

2 Answers 2

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Any vector in the plane with equation $3x+y-z=0$ since a directing vector of the line is $^{\mathrm t\mkern-1.5mu}(3, 1,-1)$.

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A line, like your $$ r = (1,1,2)^T + s (3,1,-1)^T \quad (s \in \mathbb{R}) $$ is one dimensional.

The description by a normal works via specifying the set of all vectors orthogonal to the normal and the resulting set of points is a hyperplane, a $n-1$ dimensional affine space.

So for $n=3$ that hyperplane is $3-1= 2$ dimensional, an affine plane.

For $n=2$ it is $2-1$ dimensional, thus a line. So here, in 2D, we can speak of a normal vector of a line