Equations of lines and planes.
Level 7 in this video.
We're going to go over slightly more challenging examples involving the scalar equation of a plane.
Let's jump straight into the first example.
Find these scalar equation of a plane through Point B and parallel to the plane in this problem.
We are given the coordinates of a points and a 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 points.
Any 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 these scalar equation.
Other plane parallel to it.
So we take a look at the coefficients of a 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.
All right let's take a look at the next example find a 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 a vector normal to the plane to write these 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's
sake let's set it equal to zero setting the parameter equal to zero.
We obtained the following coordinates for the points.
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 spitting the coordinates of the points 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.
Finally scalar equation of a plane that passes through the points a b and c in this problem we are given
the coordinates of three points and are asked to determine the scalar equation of a plane that contains
these three points as usual.
We need eight points.
Any vector that is normal to the plane to write these scalar equation of the plane.
We already know three points that are on the plane.
So we got plenty of points to use for our initial points.
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 three non co linear points.
These points define a plane in space.
We can generate two separate non parallel vectors from these three 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 a 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 code factors.
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 points.
So we substitute the coordinates a 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 a scalar equation of a plane that passes through point p and contains
line L in this problem we are given the coordinates of points and parametric equations of a line.
We are asked to determine the scalar equation i.e. plane that passes through point p and contains line
l similar to the previous problems.
We need a point on the plane in 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 find two non parallel vectors that are on the plane so that we can cross them in obtaining
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 points and use point P as a terminal point.
We first need to check and make sure that point p is not on the line says we want our vectors to be
non parallel.
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 defined a second vector with an initial point on the line and terminal point p.
Let's use the point on the line with a parameter t equals zero 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 non parallel vectors on the plane.
Let's go ahead and cross them so we can set up a determinant and expand by co factors.
Obtaining the following vector this vector is normal to both vector V and vector w as well as the plane.
Now it is just a matter of substituting the coordinates of point p in 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.
All right.
In our next video we will go over how to determine the angle and the equation of a line between two
intersecting planes.