Given normal to plane and a point in it, find a unit vector in plane

Question

I have a plane for which the unit normal vector and a point in the plane are known. I want to find a unit vector lying in the plane (any one). While there are dirty ways of doing this, I remember from high school that there used to be an easy way of doing this. Can someone help?

Answer 1

Consider two points on the place A$(x_1,y_1,z_1)$ and B$(x,y,z)$.
With the constraints you have, you have limited yourself to a set of vectors than a single vector, i.e., you will have parameter let's assume $"t,d"$ which is going to vary independently. The parameter $"t,d"$ will be obtained from the variable assumed by B$(x,y,z)$ .
(By saying that you want all the vector in the plane is similar to saying you want the plane itself and hence two parameters, similar to like one goes in lateral and other one in longitudinal direction of plane.)

This type of form of plane is known as one point form.

Now there can be two ways by which you can define the vector with parameter.

1) By the property of plane that $\vec r.\hat{n}=0$, where $\vec r$ lies in the plane.

$\vec r=(x-x_1)\hat i+(y-y_1)\hat j+(z-z_1)\hat k$ ,$\;\;\;\;\;$ $\hat n=a\hat i+b\hat j+c\hat k$ and $\sqrt{a^2+b^2+c^2}=1$

Note that you are assuming, $x_1,y_1,z_1,a,b,c$ are all constant and specified.
Let $x=t$ and $y=d$ (parameters)
Then, $z=\large{\frac{ax_1+by_1+cz_1-at-bd}{c}}$

Putting in $\vec r$
$\vec r=(t-x_1)\hat i+(d-y_2)\hat j+(\large{\frac{ax_1+by_1-at-bd}{c})}\hat k$

Now you can set $t,d$ anything you like you will get a vector lying in the plane.
2) By the fact that , if we assume $\vec b=\vec r\times\hat n$ then, $\vec b$ is also going to lie in the plane just perpendicular to both $\vec r$ and $\hat n$ .
I leave this one for you to try :)