# Planes

• ### Definition of Plane

Let there be three non-collinear points. There exist 3 distinct lines, which pass through these points taken 2 at a time. The triangle so formed is a planar surface.

A line joining any 2 points on a plane always lies on the plane.

• ### Equation of Plane : Normal Form

Consider a plane. Let ${N}$ be a point on this plane, such that. Position vector of ${N}$, ${\vec n}$ would be ${p \hat n}$, if ${p}$ is the magnitude of ${\vec {ON}}$. ${\hat n}$ is the unique vector along ${\vec {ON}}$. Let ${P (\vec r)}$ be any point on the plane. Clearly, ${\vec {ON} \perp \vec {NP} \ i.e. \ \hat n \perp \vec {NP}}$.

Therefore,

${(\vec r - p \hat n) \cdot \hat n =0 \ i.e. \ \vec r \cdot \hat n = p}$

Cartesian Equivalent of normal form is ${ax+by+cz+d=0}$, where ${a,b}$ and ${c}$ are the direction ratios of any line normal to the plane.

• ### Equation of Plane through a Point and Normal to a Vector

Let ${A (\vec a)}$ be the point and ${\vec b}$ be the vector. If ${\vec b}$ is perpendicular to the plane, it will be perpendicular to any vector in that plane. Let ${P (\vec r)}$ be a point on the plane. Then,

${\vec {NP} \perp \vec b \ i.e. \ (\vec {NP}) \cdot \vec b =0 \ i.e. \ \vec r \cdot \vec n - \vec a \cdot n = 0}$

Cartesian Equivalent of above equation is

${(x-a_x)b_1 + (y-a_y)b_2 + (z-a_a)b_3 =0,}$

if ${\vec a = a_x \hat i + a_y \hat j + a_z \hat k}$ and ${\vec b = b_1 \hat i + b_2 \hat j + b_3 \hat k}$.

• ### Equation of Plane through a Point and Parallel to 2 Non-parallel Non-zero Vectors

Let ${A (\vec a)}$ be the point and ${\vec b}$ and ${\vec c}$ be the vectors. ${\vec b \times \vec c}$ will perpendicular to the plane, hence,

${(\vec r - \vec a) \cdot (\vec b \times \vec c) = [\vec r- \vec a \ \vec b \ \vec c] = 0}$

• ### Equation of Plane through 3 Non-collinear Points

Let ${A (\vec a), \ B (\vec b)}$ and ${C (\vec c)}$ be the points and ${P (\vec r)}$ be any point on the plane.Thus, ${\vec {AP}}$, ${\vec {AB}}$ and ${\vec {AC}}$ will be coplanar. Thus,

${(\vec {AP}) \cdot (\vec {AB} \times \vec {AC}) = 0}$

This is the required equation.

• ### Equation of Plane through Intersection of 2 Planes

This is similar to the equation of family of straight lines through a point (in 2D).

Note that intersection of 2 planes gives us a straight line and infinitely many planes can pass through the intersection of 2 planes. Let the planes be ${\vec r \cdot \vec n_1 = p_1}$ and ${\vec r \cdot \vec n_2 = p_2}$. The vector equation of a plane through the intersection of these 2 is given by

${\vec r \cdot (\vec n_1 + \lambda \vec n_2) = p_1 + \lambda p_2}$

• ### Angle between Two Planes

Angle between 2 planes is the angle between their normals. So, if the normals are ${\vec n_1}$ and ${\vec {n_2}}$, then the angle ${\theta}$ will be

${cos (\theta) = \frac {\vec n_1 \cdot \vec n_2}{|\vec n_1| |\vec n_2|}}$

For acute angle, we take the modulus of the RHS.

If ${cos (\theta)}$ is ${0}$, planes are perpendicular. If ${cos (\theta)}$ is ${\pm 1}$, planes are either parallel or coincident.

• ### Angle between a Line and a Plane

Let the line be ${\vec r = \vec a + \lambda \vec b}$. Let ${\vec n}$ be a vector normal to the plane. Clearly, angle between the line and the plane will be equal to ${90^o}$ minus the acute angle made by the line (or ${\vec b}$) and ${\vec n}$.

• ### When are 2 Lines Coplanar?

We know that in 3 dimensions, lines can be skew or coplanar. Let the lines be ${\vec r = \vec a_1 + \lambda \vec b_1}$ and ${\vec r = \vec a_2 + \mu \lambda b_2}$ be the lines. The lines will be coplanar if and only if the scalar triple product of the following vectors is zero :

${(\vec a_2 - \vec a_1), \ \vec b_1, \ \vec b_2}$

• ### Distance of a Point from a Plane

This is similar to distance of a point from a line. We find the value of parameter ${lambda}$. Let ${A (\vec a)}$ be the point and ${\vec r \cdot \vec n = p}$ be the plane. Let the foot of the perpendicular from ${A}$ on the plane be ${M}$. Vector equation of line ${AM}$ will be

${\vec r = \vec a + \lambda \vec n}$

Since ${M}$ lies on the plane, we have

${(\vec a + \lambda \vec n) \cdot \vec n = p}$

From this, we get the value of ${\lambda}$. Having obtained ${\lambda}$, we can easily get coordinates of ${M}$ and eventually the distance ${AM}$.