Planes

Equation of planes can expressed in either vector or cartesian form. Vector equation is the one containing only vectors. Cartesian equation is in the form: $Ax+By+Cz=D$.

Contains a Point and Parallel to 2 Vectors

Suppose a plane:

• is parallel to both $\underline{a}$ and $\underline{b}$ where $a\times b \neq 0$
• contains $\underline{r_0}=x_0\underline{i}+y_0\underline{j}+z_0\underline{k}$

Equation for the plane is:

Contains a Point and Normal Is Given

Suppose a plane:

• contains $\underline{r_0}=x_0\underline{i}+y_0\underline{j}+z_0\underline{k}$
• has a normal $\underline{n}$

Equation for the plane is:

Contains 3 Points

Suppose a plane contains $r_0,r_1,r_2$ ($\underline{r_0},\underline{r_1},\underline{r_2}$ are the position vectors of respectively).

Normal to a Plane

Suppose $ax+by+cz=d$ is a plane. $\underline{n}=a\underline{i}+b\underline{j}+c\underline{k}$ is a normal to the plane.

Angle Between 2 Planes

Consider the two planes:

• $A: a_1x+a_2y+a_3z=d$
• $B: b_1x+b_2y+b_3z=d'$

The angle between the planes $\phi$ is given by:

Here $\underline{n_A},\underline{n_B}$ are normal to the planes $A,B$.

Shortest Distance from a Point

Consider the plane $ax+by+cz=d$.

• $\underline{n}$ is a normal to the plane
• $\underline{r_0}$ is the position vector of any known point on the plane
• $\overline{r_1}$ is the position vector to the arbitrary point

Intersection

In 3D, to prove 2 planes intersect, it has to be proven that there is a point satisfiying both of the planes.

Of 2 Planes

Can either be a:

• Plane - when the planes coincicde
• Line - otherwise

Equation of the line of intersection can be found by:

• Solving $y,z$ with respect to $x$
• Subject $x$ and symmetric form can be found

Of 3 Planes

Can either be a:

• Plane - when the planes coincide
• Line - when the lines of intersection between the planes pairwise coincide
• Point - otherwise

First pairwise intersection of the planes must be found. And then intersection of those 2 can be found.