# Given a vector, how to compute orthogonal plane

1. Oct 9, 2015

### MF2

Given a vector (in 3-d), how do I determine the plane that is orthogonal to it?

I am not quite finding a search term that gets me to this, but instead to several similar, but different questions.

One such is find an equation of a plane perpendicular to a vector and passing through a given point. I may have been staring at this too long, but I am not visualizing why the given point is needed.

Thanks!

2. Oct 9, 2015

### Daeho Ro

When the plane is given by
$$a x + b y + c z + d = 0,$$
the normal vector of that plane is $(a,b,c).$ Then, you just replace the normal vector to the vector what you want and find $d$ by using any point on the plane.

3. Oct 9, 2015

### Jack Davies

There are infinitely many planes orthogonal to a given vector, so you would also need to specify a point on the plane to calculate its equation. You can write a plane with normal $\bf{n}$ in vector notation as the set of all $\bf{x}$ such that $$\bf{x \: . n} = d$$ Where $d$ is a scalar determined by a given point. Alternatively you can expand this out in cartesian coordinates as Daeho Ro wrote above.

4. Oct 12, 2015

### MF2

The plane would go through one end of the vector.

For example, a 3-d vector would go from point (0,0,0) to (x,y,z). I want the plane perpendicular to this vector, going through (x,y,z). Do I need another point to define the plane, or is this enough information.

Thanks!

5. Oct 12, 2015

### Staff: Mentor

To get the equation of a plane, you need a normal to the plane, and two points on the plane, one of which is arbitrary and has coordinates (x, y, z). Call these points P(x, y, z) and $P_0(x_0, y_0, z_0)$, where the coordinates $x_0, y_0, z_0$ are known.
Form a direction vector $\vec{v} = <x - x_0, y - y_0, z - z_0>$. The normal and this vector are perpendicular, so their dot product $\vec{v} \cdot \vec{N} = 0$. This dot product will produce the equation of the plane.