Question about the formula of lines and planes

Question

I just watched some videos from Khan Academy which left me slightly confused. I don't really understand why the formula of a plane is $n(p_1-p_2) = 0$, where $n$, $p_1$ and $p_2$ are vectors.

I do understand the part that $n$ is a vector orthogonal to the plane and thus also orthogonal to the vector $p_1-p$, and that the dot product of those two vectors is $0$.

The part I don't understand is, why should I take $n(p_1-p_2) = 0$ and not $n*(p_1-p_2) + p_1 = 0$.

See the drawing below. In the plane, the orange vector is then $p_1 - p_2$, but in the two dimensional vector I would have to add $p_1$ to it (if I understood correctly) otherwise I would get the vector on the bottom (of the first drawing). So why shouldn't I do this in the three dimensional vector? I must have misunderstood something or missed something.

Answer 1

Yes, you did misunderstand this thing!

Let's go by your logic. You are saying $p_1 - p_2$ denotes this bottom vector, for the moment, say we ignore the vector translation property. Now, denote by $c$ the vector between the tips of $p_1 - p_2$ with the direction being from $p_1$ tip to $p_2$ tip.

Now $p_1+c = p_2 => p_1-p_2=-c$

So $c$ lies in the top orange line, though direction reversed.

Answer 2

Your analogy in the euclidean plane is erroneous I'm afraid. Let $P_0$ be a given point and $P$ be an arbitrary point in the plane, $n$ be a normal vector to the plane, and denote by $v_0:=\vec{0P_0}$ $v:=\vec{0P}$ the vectors from the origin to $P_0$ and $P$, respectively. Then $v-v_0$ is a vector lying in the plane, and as we can keep track of any point in the plane in terms of this vector, $n\cdot(r-r_0)=0$ gives the equation to the plane, i.e. the plane is the set $\{P\in\mathbb{R}^n|n\cdot(r-r_0)=0\}$. The intuitive idea behind this is that two points on the plane and one direction to face determines the plane.

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The analogy of this idea in the euclidean plane would be as follows: We have a line we want to parametrize (which is your orange "vector"). Note that we would not need the notion of a normal to define a line. Using the same notation as before, let $P_0,P_1$ be two points on the line and $v_0,v_1$ be the corresponding vectors. Then $v_1-v_0$ is a vector lying on the line, and consequently the line is given by $\{Q\in\mathbb{R}^2|\exists t\in\mathbb{R}:Q=P_0+t(v_1-v_1)\}$. The intuitive idea behind this is that two points determine a line (taking into consideration the intuitive idea of the equation of the plane given above, for a plane we first determine a line on the plane, and then not let the plane tilt on the line).

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In your picture you take vectors $p_1,p_2$ that do not lie on the line you want to write an equation for, and the same goes for your space picture. So the moral is that it may be wise to keep track of vectors and points.