Equations of lines and planes.
Level 6.
In this video we're 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 into non parallel
direction vectors that start at the same point on the plane.
Now what makes the equations of airline and plane in space different from each other is that the equation
of a line is set to have one 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 Tripoli quality 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 coordinates equal to 9.
This forces the two other coordinates to obtain the following values on the other hand.
The equation of a plane in space has two degrees of freedom so we can randomly choose a value for two
variables and we can determine the value of a third variable.
We can essentially choose two of the three unknowns at random and saw for the third unknown in this
case choosing two random values for each of the two chosen variables will force you to choose the value
of a third variable.
For example in the following scalar equation of a plane if we let x equal to and Y equal to that 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 this thing to x and y coordinates.
But each combination of x and y coordinates has a unique value for the z coordinates.
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 or a set of direction numbers for a line or plane then came a K B and KC where k is an
element of the real numbers and K does not equal 0 is also a sense of direction numbers for that line
or plane in space this means that the equation of lines and planes or not unique also the initial point
x 0 y 0 and 0 is also not unique.
There are infinite choices for this initial point for both airline and plane in space.
All right.
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 points 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 Skitter equation will not be satisfied.
Okay so let's go ahead and substitute the coordinates of the first points 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 points 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 all right
let's go over the next example find the equation of the plane to the point A and perpendicular to vector
V in this problem.
We are given the coordinates of a points 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 Skeeter 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 appoint a and the components of vector V into these
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.
All right.
Let's take a look at the next example find parametric equations for the line three point p that is perpendicular
to the plane.
In this problem we are given the coordinates a 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 points any 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 points 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 points 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 three point p and perpendicular to the line in this problem.
We are given the coordinates of point p in the parametric equations of a line.
We are asked to find the scalar equations 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 in 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.
Any normal vector.
Let's go ahead and substitute them into the scalar equation of a plane.
So we substitute the coordinates a 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 or final answer all right.
In our next video we will go over slightly more challenging examples involving the equation of a plane
in space.