welcome to the equation of a plane using the normal vector
the standard form of an equation of a plane
contain the points x sub one y sub one z sub one
and having the normal vector with components a,b and c
is given by a times the quantity x minus x one
plus b times the quantity y minus y one
plus c times the quantity z minus
z one equals zero so the point on the plane
give us the constants of
the binomial factors and the components of the normal vector abc
give us the constant factors
a,b and c
now once we have this equation in standard form we can distribute
combine like terms to come up with what's called the general form
of a plane let's take a look at the idea behind
where this equation comes from here's the graph a plane
and the normal vector in black now if we pick any two points in the plane to form
another vector
it would still have to be perpendicular or or orthogonal to the
normal vector so we were to find the dot product
of the normal vector in any vector in this plane
would have to equal zero be because they would be orthogonal
and this is the idea where the equation the plane comes from
so again if we take the component reformer the normal vector
which is perpendicular to plane and dotted with
another vector form by any two points in the plane
the second vector would have component form as we see here
and since this vector is in the plane
and the vector is orthogonal to the plane the dot product
must equal zero and this dot product gives us the equation of
that plane let's go and take a look at an example
here we want the standard and general form
of an equation of a plane that contains the point
three negative two five and has a normal vector
five two negative three so the equation will be five
times x minus three plus
two times y
minus negative two or y plus two and the z component of the normal vector is negative three so
we'll write down minus three times
z minus 5 and this must
equal zero and this is the standard form
of the plane again to emphasize where this equation came from
the components of the normal vector gave us the five
the two and the negative three and then the point
three negative two five gave us the constants
the binomial factors now to find the general equation
we just need to distribute and collect like terms
so we have five x minus fifteen
plus two y plus four
minus three z plus fifteen
equals zero so were gonna have five x
plus two y minus three z
and our constant term is going to be negative fifteen
plus four plus fifteen that's gonna give us positive four
equals zero let's go and take a look at the graph of this plane
in the next we will take a look at how we can go about graphing these on our own paper
but for right now if we rotate the coordinate system you get a nice
idea of where this plane is in space
let's go and take a look at another example we want to determined the standard
and general equation of a plane containing
these three points well this problem can take a little bit more work
we do have to determine the normal vector to this plane
one way to do that is to determine two vectors in the plane
contain these two points in component form
we'll find their cross product which will give us their normal vector
so lets first find the component form of two vectors
using the street points we'll call this point p
this point q and this point r
lets find the component form of vector pq
that'll be negative three minus one
four minus negative one or four plus one
and negative one minus two
so the component form will be negative four
five negative three
now let's find the component form a vector pr
will have three minus one
negative two minus negative one or negative two plus one
and five minus two
so the component form of the second vector will be two
negative one and three
so again these are two vectors in the plane so if we cross these two vectors we
will get the normal vector
which will help us find the equation that plane
so lets just go ahead and call this vector v
call this vector w and we are going to cross the two vectors
remember that going to be equal to the determinant
where the first row be i,j,k
the second row will be negative four five negative three
in the third row will be two negative one three
lets go and evaluate using the co-factor expansion method
this is going to be equal to the two by two determinant formed by eliminating row one
and column one so we'll have five negative 3
negative one three
times vector i minus
the two by two determinant formed by eliminating row one in column two
so we'll have negative four negative three
two three times
vector j plus
the two by two determinant formed by eliminating row one in column 3 so we'll have
negative four five
to negative one times vector k
so we're gonna have fifteen minus
positive three that'll be twelve i
minus it's going to be negative twelve
minus negative six or negative twelve plus six, that will be negative six
minus negative six will be plus six
j plus
here were going to have four minus ten that's going to be negative six
so we'll change this to minus six k
so the component form of the normal vector
is going to be twelve six
and negative six now any normal vector will work
so I could take a scalar multiple of this normal vector
instead of using this one and I think I'll do that because
all of these components have a common factor of six
so we'll use one sixth times this vector
as our normal vector so we're gonna have
two one negative one for our normal vector
let's go ahead and take that over the next screen
now we have to decide which point we want to use to create the equation of the plane
let's go back to the previous slide let's go ahead and use this first point p
one negative one two
this is all we need to find the standard equation of this plane
we're gonna have two times x minus one
plus one times y
minus negative one or y plus one
minus one times z minus two
and this must equal zero
now we go ahead eliminate these ones here if we wanted to
but we'll go ahead and call this the standard equation the plane
and now to find the general equation we'll go ahead and just distribute
and combine our like terms so we'll have two x minus two
plus y plus one minus
z plus two
equal zero, so we'll have two x
plus y minus z
and our constant term is going to be negative two
plus one plus two that will be plus one
and this equals zero, that will do it for this video
thank you for watching