Equations of Lines and Planes Level 7.
In this video we are going to go over slightly more challenging examples involving the scalar
equation of a plane.
Let’s jump straight into the first example.
Find the scalar equation of a plane through point B and parallel to the plane.
In this problem we are given the coordinates of a point and the scalar equation of a plane.
We are asked to find the scalar equation of a plane that passes through point B and is
parallel to the plane.
In order to find the scalar equation of a plane we need a point and a vector normal
to the plane.
We already know that both planes are parallel to each other so both planes will also have
the same normal vector.
We can determine the normal vector of the plane by using the scalar equation of the
plane parallel to it; so we take a look at the coefficients of the scalar equation of
the plane and obtain the following normal vector.
Now it is just a matter of using this normal vector along with point B to write the equation
of the plane.
So we substitute the coordinates of point B and the components of the normal vector,
then we simplify the expression, doing that we obtain the following scalar equation of
the plane and this is our final answer.
Alright let’s take a look at the next example.
Find the scalar equation of a plane that contains line L and is parallel to the plane.
In this problem we are given the parametric equations of a line and the scalar equation
of a plane.
We are asked to determine the scalar equation of a plane that contains line L and is parallel
to the plane.
Similar to the previous example we need a point on the plane and vector normal to the
plane to write the scalar equation of the plane.
Since the plane contains this line we can determine a point on the plane by using the
parametric equations of line L, since any point on this line will also be on the plane.
We can easily generate a point on the line by setting the parameter equal to a value,
for simplicity sake let’s set it equal to zero.
Setting the parameter equal to zero we obtain the following coordinates for the point.
Now we need to determine a vector normal to the plane, we will obtain the components of
this vector by using the coefficients of the scalar equation of the parallel plane.
We could use these components for the normal vector but remember that any constant multiple
of this vector will also work, so let’s divide these components by two to obtain nicer
components.
Now it is just a matter of substituting the coordinates of the point and the components
of the normal vector.
Then we go ahead and simplify the expression and move the constant to the other side, doing
that we obtain the following scalar equation of the plane and this is our final answer.
Let’s move along to the next example.
Find the scalar equation of a plane that passes through the points A, B and C.
In this problem we are given the coordinates of 3 points and are asked to determine the
scalar equation of a plane that contains these 3 points.
As usual we need a point and a vector that is normal to the plane to write the scalar
equation of the plane.
We already know 3 points that are on the plane so we got plenty of points to use for our
initial point.
What we don’t have is a normal vector to the plane.
We need to figure out a way to determine a normal vector.
Notice that we have 3 non-collinear points; these points define a plane in space.
We can generate two separate nonparallel vectors from these 3 points and then we can use the
cross product to find a vector that is perpendicular to both of these vectors.
The vector that is obtained by crossing these two vectors can be used as the normal vector
for the plane.
So let’s define one vector with initial point A and terminal point B as vector V,
and a second vector with initial point A and terminal point C as vector W.
Next let’s subtract the corresponding coordinates of the initial point with the corresponding
coordinates of the terminal point for each vector.
Doing that we obtain the following components for vector V and vector W. Now we need to
obtain a vector that is perpendicular to both of these vectors so we go ahead and cross
these vectors.
Setting up the determinant and expanding by cofactors we obtain the following vector.
This vector is orthogonal to both vector V and vector W as well as the plane that contains
all three points.
The last step is to substitute the coordinates of one of these points and the components
of the normal vector.
Let’s use point A for our initial point, so we substitute the coordinates of point
A and the components of the normal vector into the scalar equation of the plane.
Simplifying and moving the constant to the other side we obtain the following scalar
equation of the plane and this is our final answer.
Let’s go over the final example.
Find the scalar equation of a plane that passes through point P and contains line L.
In this problem we are given the coordinates of a point and parametric equations of a line.
We are asked to determine the scalar equation of a plane that passes through point P and
contains line L. Similar to the previous problems we need a point on the plane and a normal vector
to the plane.
We already have a point so we need to figure out how to obtain the components of a vector
normal to the plane.
We need a way to define two nonparallel vectors that are on the plane so that we can cross
them and obtain a normal vector just like we did in the previous example.
We can define one of these vectors just by inspection by using the parametric equations
of the line, since this line is on the plane the direction vector of this line will also
be on the plane, let’s denote it as vector V.
Now we need a second vector on the plane, we can create a second vector by using any
point on the line as the initial point and use point P as the terminal point.
We first need to check and make sure that point p is not on the line since we want our
vectors to be nonparallel; we determine this by using the direction vector of the line.
Point P will be on the line if there is a constant multiple of the direction vector
that generates a displacement vector with terminal point P. So we go ahead and solve
the following vector equation by breaking it down into components and solving for the
constant.
Notice that we obtain different values for the constant so we now know that point P is
not on the line.
We can now define a second vector with an initial point on the line and terminal point
P.
Let’s use the point on the line when the parameter t equals 0 for our initial point.
Subtracting the corresponding coordinates of the initial point from the corresponding
coordinates of point P we obtain the following components of the vector, let’s denote it
as vector W. Now that we have two nonparallel vectors on the plane, let’s go ahead and
cross them, so we can set up a determinant and expand by cofactors obtaining the following
vector.
This vector is normal two both vector V and vector W as well as the plane.
Now it is just a matter of substituting the coordinates of point P and the components
of the normal vector into the scalar equation of the plane, doing that and simplifying we
obtain the following scalar equation of the plane and this is our final answer.
Alright in our next video we will go over how to determine the angle and the equation
of a line between two intersecting planes.