Why is the perpendicular vector on a plane always the vector of coefficients of the variables in the plane equation?

e.g., for the plane $2x-y+3z=8$, the perpendicular vector is $(2,-1,3)$.


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    $\begingroup$ Consider dot product $\endgroup$ Apr 23 '15 at 19:11

Take two points on the plane: $(x_1,y_1,z_1),(x_2,y_2,z_2)$. Then they both satisfy the plane equation: $$2x_1-y_1+3z_1=8,$$ $$2x_2-y_2+3z_2=8.$$This gives $\left < x_1-x_2,y_1-y_2,z_1-z_2 \right > \cdot \left < 2,-1,3 \right > =0$. In other words, any vector on the plane is perpendicular to the vector $\left < 2,-1,3 \right >$.

  • $\begingroup$ Why does this work? The proof proves it, but I do not understand why. $\endgroup$
    – Tsangares
    Feb 6 '17 at 2:31
  • $\begingroup$ @Tsangares, can you specify which part you do not understand? $\endgroup$
    – Easy
    Feb 7 '17 at 4:25
  • $\begingroup$ I don't understand why x1-x2 is on the plane; if we have the line b=a+1 then I can take the points x1=(1,2), x2=(2,3) that lie on the line but x1-x2=(-1,-1) doesn't lie on the same line... $\endgroup$
    – volperossa
    Dec 25 '17 at 16:11
  • $\begingroup$ @volperossa it is parallel to that, that's the description of vectors. $\endgroup$ Jun 3 '19 at 12:23
  • $\begingroup$ @volperossa you may refer to this website. $(x_i,y_i,z_i)$ is the position vector of the point, the difference of which will lie in the plane. $\endgroup$
    – Chris Tang
    Feb 5 '20 at 3:57

I am assuming we are on $\mathbb{R}^{3}$.

A plane is determined by a point on the plane and a vector orthogonal to the plane. Say $P_0$ is a point on this plane, and $\vec{n}$ is the orthogonal vector. Also let's show the position of point $P_0$ with the vector $\vec{r_{0}}$.

Now take a generic point on the plane, call it $P$, show its position with the vector $\vec{r}$. Then the following equation has to be satisfied:

$\vec{n}.(\vec{r} - \vec{r_{0}}) = 0$

This has to hold because the difference vector $\vec{r} - \vec{r_{0}}$ has to lie in our plane.

Now let's try to find out the scalar equation for our plane. First start by substituting our vectors:

$\vec{n} = \langle a, b, c \rangle$, $\vec{r} = \langle x, y,z \rangle$ and $\vec{r_{0}} = \langle x_{0}, y_0, z_0 \rangle$.

Expanding our vector multiplication $\vec{n}.(\vec{r} - \vec{r_{0}}) = 0$ will give us:

$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$

If you define $d = x_0 + y_0 + z_0$, then the previous equation becomes:

$ax + by + cz = d$.

From this equation you can identify our normal vector $\vec{n}$ directly from the coefficients $\langle a, b, c \rangle $ of the scalar equation of plane.

  • $\begingroup$ I don't understand why (r-r0) has to lie on the plane; for example: if we have the line y=x+1 then I can take the points r=(1,2), r0=(2,3) that lie on the line but r-r0=(-1,-1) doesn't lie on the same line... $\endgroup$
    – volperossa
    Dec 25 '17 at 16:04
  • $\begingroup$ @volperossa, (r-ro) is the vector subtraction and not the difference in the magnitudes in two dimension. The vector (r-ro) corresponds to the line segment joining r and r0 along the line y=x+1 $\endgroup$
    – gaganso
    Oct 7 '18 at 16:58
  • $\begingroup$ Is not The vector subtraction as vector sums i.e. The difference of components? $\endgroup$
    – volperossa
    Oct 7 '18 at 17:04

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