I have come across this question that I need a tip for.

Find the equation (general form) of the plane passing through the point $P(3,1,6)$ that is orthogonal to the vector $v=(1,7,-2)$.

I would be able to do this if it said "parallel to the vector"

I would set the equation up as $(x,y,x) = (3,1,6) + t(1,7,-2)$ and go from there.

I don't get where I can get an orthogonal vector. Normally when I am finding an orthogonal vector I have two other vectors and do the cross product to find it.

I am thinking somehow I have to get three points on the plane, but I'm not sure how to go about doing that.

Any pointers?

thanks in advance.

  • 3
    $\begingroup$ $(x,y,x) = (3,1,6) + t(1,7,-2)$ is not a plane, it is a line.... $\endgroup$
    – N. S.
    Nov 14, 2011 at 22:07

4 Answers 4


For a plane in $\mathbb{R}^3$ with $\mathbf{r_0}$ a point that lies in the plane and $\mathbf{n}$ a vector normal to the plane, its equation can be given by: $$ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad \mbox{where} \quad \mathbf{r}=(x,y,z) $$

To solve your question, let us first rearrange the above equation: $$ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}-\mathbf{n}\cdot\mathbf{r_0}=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}=\mathbf{n}\cdot\mathbf{r_0} $$

Substituting values, gives the following equation for the plane in Cartesian form: $$ (1,7,-2)\cdot(x,y,z) = (1,7,-2)\cdot(3,1,6) \quad \Rightarrow \quad x+7y-2z = -2 $$

  • $\begingroup$ @Cheesegraterr: If you would like more details, please let me know :) $\endgroup$
    – matt
    Nov 27, 2011 at 6:13
  • $\begingroup$ Does anything change if n is (or isn't) unit length? $\endgroup$ Feb 11, 2016 at 17:38

Key point:

A plane with normal vector $(n_1, n_2, n_3)$ has equation $$n_1 x + n_2 y + n_3 z = k$$ for some $k$.


Try solving $(P-r)\cdot v=0$, where $r =(x,y,z)$.


vector equation of a plane is in the form : r.n=a.n, in this case, a=(3,1,6), n=(1,7,-2). Therefore, r.(1,7,-2)=(3,1,6).(1,7,-2)






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