Suppose we have the plane with equation $3x-7z=12$. How to find its normal vector?

The plane with equation $Ax+By+Cz+D=0$ has the normal vector $\mathbb{n}=(A,B,C)$. Using this we get that above plane has normal vector $(3,0,-7)$, right?

Let's apply another method. Take three points which lies on the plane, namely $A=(4,0,0), \ B=(0,0,-\frac{12}{7}), \ C=(1,0,\frac{-9}{7})$ then vector $AB=(-4,0,-\frac{12}{7})$ and $AC=(-3,0,-\frac{9}{7})$. Taking their cross product we get zero vector. What is wrong in my idea?

  • $\begingroup$ You have an error: $AC=(-3,0,9/7)$. $\endgroup$ Oct 14, 2017 at 17:00
  • $\begingroup$ The third component of point $C$ shoul have sign minus, namely $-9/7$ $\endgroup$
    – RFZ
    Oct 14, 2017 at 17:05
  • $\begingroup$ I don't think that's true. $\endgroup$ Oct 14, 2017 at 17:06
  • $\begingroup$ Why? Maybe I am tired of. Could you show it? $\endgroup$
    – RFZ
    Oct 14, 2017 at 17:12
  • $\begingroup$ I'll add an answer. $\endgroup$ Oct 14, 2017 at 17:12

4 Answers 4


In response to the first part:

Suppose two points, $P(x,y,z)$ and $P_0(x_0, y_0, z_0)$ lie on a plane with a normal vector $\mathbf{n}$. Also, let $\mathbf{r}$ denote the position vector of $P$ and $\mathbf{r_0}$ denote the position vector of $P_0$.

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The picture clearly shows that $\mathbf{n}\cdot(\mathbf{r-r_0}) = 0$, since the two vectors are perpendicular. Then, $$\mathbf{n}\cdot \mathbf{r} = \mathbf{n}\cdot\mathbf{r_0}$$ In your case, we have $3x+0y-7z = 12$, which is equivalent to $(3,0,-7)\cdot(x,y,z) = 12$. Comparing with above, we have the components of the normal to the plane are $(3,0,-7)$.

Second question:

You have made a computation mistake, as has been pointed out in the other posts. Your idea is nonetheless true: if you have two direction vectors lying in the plane, then their cross product will result in a vector orthogonal to both these vectors, i.e. normal to the plane.


Your approach is valid, as indeed computing the cross product of two vectors $v,w\in \mathbf{R}^3$ yields a resultant vector $v\times w\in \mathbf{R}^3$ orthogonal to the original vectors. Now, you have fixed points $A=(4,0,0),\:B=(0,0,-\frac{12}{7}),\:C=(1,0,\frac{9}{7})$ in the plane. We can construct vectors that lie in the plane by defining $\vec{AB}= (B-A), \vec{AC}=(C-A).$

Computing component by component yields

$$\vec{AB}=(0-4,0-0,-\frac{12}{7}-0)=(-4,0,-\frac{12}{7})$$ $$ \vec{AC}=(1-4,0-0,\frac{9}{7}-0)=(-3,0,\frac{9}{7}).$$ Computing the cross product of these vectors yields a resultant vector orthogonal to the plane. Your error is that you made a subtraction mistake, and ended up with vectors $v=(-4,0,-\frac{12}{7}), w=(-3,0,-\frac{9}{7})$. Note that $\frac{4}{3}w=v$. So, because these vectors are colinear, their cross product is degenerate and returns the $0$ vector.

  • 1
    $\begingroup$ We have an equation $3x-7z=12$ I don't think that point $C=(1,0,9/7)$ lie on plane because if we put them into plane's equation we don't get identity. Since $3*1-7*\frac{9}{7}=-6\neq 12$ $\endgroup$
    – RFZ
    Oct 14, 2017 at 17:24
  • $\begingroup$ @ZFR agree. That's what I was thinking. $\endgroup$
    – Milan
    Aug 8, 2020 at 0:19

the vector $$\vec{AC}$$ is given by $$\vec{AC}=\left(-3;0;\frac{9}{7}\right)$$ $$A(4;0;0),C\left(1;0;\frac{9}{7}\right)$$ then $$\vec{AC}=\left(1-4;0-0;\frac{9}{7}-0\right)$$

  • $\begingroup$ Your point $C$ does not lie on plane $3x-7z=12$ since $3*1-7*\frac{9}{7}=-6\neq 12$ $\endgroup$
    – RFZ
    Oct 14, 2017 at 16:59

Unfortunately, your two chosen vectors are parallel (all three points are on a line in the plane) so their cross-product will necessarily be zero; if you choose one of your three points to not lie in the same line in the plane then the cross-product will be parallel to your extracted normal vector.


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