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}.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s