Equations of Lines and Planes.
Level 6.
In this video we are going to compare the equations of lines and planes and then we
will go over a couple of basic examples.
Before we go over a couple of examples it is important to compare the equations of lines
and planes since they both have very similar expressions and they can be easily mixed up.
Let’s first quickly review how to describe a line in space.
The first way to describe a line in space is by using either a vector or parametric
equations; these equations require you to know an initial point on the line and a direction
vector parallel to the line.
The second way is by using the symmetric equations.
The third way is used to describe a line segment in space which uses a two-point vector equation
with a parameter that ranges from 0 to 1 inclusive.
There is also three ways to describe a plane in space.
The first way to describe a plane in space is by using a vector equation; this equation
requires a point on the plane and a vector normal to the plane.
The second way is by using the scalar equation which also requires a point on the plane and
a vector normal to the plane.
The third way is by using parametric equations which require a point on the plane and two
non-parallel direction vectors that start at the same point on the plane.
Now what makes the equations of a line and plane in space different from each other is
that the equation of a line is said to have 1 degree of freedom.
Notice that the symmetric equations of a line in space is written as a triple equality,
we have the freedom to randomly choose a value for one of the variables either x,y, or z
and we can find the value of the other two variables (since the triple equality forces
you to choose the two other values, once you choose a value for one of the variables it
will fix the value of the other two variables).
For example in the following symmetric equation of a line if we let x equal 9 then we can
figure out the value for y and z.
If you think about this geometrically by visualizing a line cutting through space notice that there
is exactly one point on the line that will have an x coordinate equal to 9, this forces
the two other coordinates to attain the following values.
On the other hand the equation of a plane in space has 2 degrees of freedom so we can
randomly choose a value for two variables and we can determine the value of the third
variable.
We can essentially choose two of the 3 unknowns at random and solve for the third unknown.
In this case choosing two random values for each of the two chosen variables will force
you two choose the value of the third variable.
For example in the following scalar equation of a plane if we let x equal 2 and y equal
2 then we can solve for the value of z.
If we visualize a plane in space notice that there are various points on the plane with
distinct x and y coordinates but each combination of x and y coordinates has a unique value
for the z coordinate.
This method of randomly choosing values for the variables based on the degrees of freedom
is the quickest way to find a point that is on the line or plane in space and will be
very useful in solving particular types of problems.
Keep in mind that the direction numbers a, b and c for a line or plane are not unique.
If a, b, and c are a set of direction numbers for a line or plane, then ka, kb, and kc,
where k is an element of the real numbers and k does not equal 0, is also a set of direction
numbers for that line or plane in space.
This means that the equation of lines and planes are not unique.
Also the initial point (x zero, y zero, and z zero) is also not unique there are infinite
choices for this initial point for both a line and plane in space.
Alright with that said let’s go over some basic examples involving planes.
Determine whether the plane passes through each point.
In this problem we are given a scalar equation of a plane and are asked to determine if the
plane passes through the following two points.
We can determine this by just substituting the coordinates of the point into the scalar
equation of the plane, if the point is indeed on the plane then the scalar equation should
be satisfied, if the point is not on the plane then the scalar equation will not be satisfied.
Ok so let’s go ahead and substituted the coordinates of the first point into the scalar
equation of the plane, then we simplify the left side of the equation, doing that we see
that we obtain a true statement so the plane passes through this point.
Now let’s do the same with the second point and substitute it into the scalar equation.
From here it is just a matter of simplifying the expression; in this case we get a false
statement so we conclude that the plane does not pass through this point and this is our
final answer.
Alright let’s go over the next example.
Find the equation of the plane through point A and perpendicular to vector V.
In this problem we are given the coordinates of a point and the components of a vector.
We are asked to find the equation of a plane that passes through this point and is perpendicular
to the vector.
We can determine the equation of the plane by using the scalar equation of a plane.
Recall that we need a point on the plane and a vector perpendicular to the plane in order
to write the equation of the plane.
So let’s go ahead and substitute the coordinates of point A and the components of vector V
into the scalar equation of a plane.
Next it is just a matter of simplifying the equation, so we distribute the components
of the vector, and then we collect like terms, moving the constant to the right side of the
equation; doing that we obtain the following equation for the plane and this is our final
answer.
Alright let’s take a look at the next example.
Find parametric equations for the line through point P that is perpendicular to the plane.
In this problem we are given the coordinates of point P and a scalar equation of a plane.
We are asked to determine the parametric equations of a line that passes through point P and
is perpendicular to the plane.
In order to find the parametric equations for the line we need a point and a vector
parallel to the line.
Notice that if the line is perpendicular to the plane then any vector that is normal to
the plane will be parallel to the line.
So all we need is to determine the normal vector to the plane and use it as a direction
vector for the line.
We can easily find a normal vector to the plane by using its scalar equation.
Recall that a normal vector to the plane can be obtained by using the coefficients of the
scalar equation of the plane.
In this case a normal vector to the plane has the following components, we can also
use any scalar multiple of this vector as well but we will use this particular normal
vector for convenience.
Now that we have a point and a direction vector for the line we can go ahead and write the
parametric equations for the line.
So we substitute the coordinates of point P for the initial point and use the components
of the normal vector to the plane for our direction numbers.
Doing that we obtain the following parametric equations and this is our final answer.
Let’s take a look at the final example.
Find the scalar equation of a plane through point P and perpendicular to the line.
In this problem we are given the coordinates of point P and the parametric equations of
a line.
We are asked to find the scalar equation of a plane that passes through point P and is
perpendicular to the line.
In order to write the scalar equation of the plane we need a point on the plane and a vector
perpendicular to the plane.
Notice that the line is perpendicular to the plane so we can use the direction vector of
the line as the normal vector for the plane.
We can determine the components of the direction vector of the line by taking a look at the
given parametric equations; in this case the direction vector will have components equal
to the coefficients of the parameter t.
This direction vector will also be used as the normal vector for the plane.
Now that we have a point and a normal vector let’s go ahead and substitute them into
the scalar equation of a plane.
So we substitute the coordinates of point P and the components of the normal vector,
from here it is just a matter of simplifying and moving the constant to the other side.
Doing that we obtain the following scalar equation of the plane and this is our final
answer.
Alright in our next video we will go over slightly more challenging examples involving
the equation of a plane in space.