### Video Transcript

In this video, we will learn how to
find points and lines of intersection between lines and planes in 3D space.

Recall that a plane in 3D space π
three may be described by the general equation ππ₯ plus ππ¦ plus ππ§ plus π
equals zero, where π, π, π, and π are all constants. Such a plane may look something
like this on a three-dimensional graph. The plane consists of the
coordinates of all possible points π₯, π¦, π§ that satisfy the equation ππ₯ plus
ππ¦ plus ππ§ plus π equals zero. Similarly to lines in 2D space,
planes in 3D space extends infinitely in all directions. Assuming the plane is not parallel
to any of the coordinate planes, that is, the π₯π¦-, the π¦π§-, or the π₯π§-plane,
for any given value of one variable, we can always find values for the other two
variables that will solve the equation of the plane.

A planeβs orientation in space is
defined by the coefficients of the variables in its equation: π, π, and π. More precisely, the coefficients
define a vector π§ that is normal to the plane and may be given by simply reading
off the coefficients of the equation of the plane. π§ is equal to π, π, π. Any nonzero scalar multiple of this
vector, πΌπ§, is also normal to the plane. So, for example, if we have the
plane two π₯ plus three π¦ plus four π§ minus five equals zero, we can obtain one
normal vector π§ by just reading off the coefficients of the variables in the
equation. So π§ is equal to two, three,
four. But any scalar multiple of π§ is
also normal to the plane, for example, two π§ equal to four, six, eight or negative
π§ equal to negative two, negative three, negative four.

If we have two different planes
with normal vectors π§ one and π§ two, respectively, where π§ two is a scalar
multiple of π§ one, then these two planes are parallel. Any two planes that do not have
parallel normal vectors will intersect over a line in 3D space. Recall that in 2D space two
nonparallel lines will intersect at exactly one point π₯ naught, π¦ naught. This point is the unique solution
to the two equations of the two lines. We have the same number of
equations as we do unknowns and therefore one unique solution.

In 3D space, things are a little
different. Two nonparallel planes will
intersect over a straight line in 3D space. This line contains the infinite
number of points π₯, π¦, and π§ that solve the two equations of the two planes. We have one more unknown than we do
equations. Therefore, there are infinitely
many solutions to these two equations which form the line in 3D space. This line is one-dimensional and
may therefore be parameterized by one parameter. Letβs look at an example of how to
find the general equation of a line between two planes.

Find the general equation of the
line of intersection between the planes π₯ minus four π¦ plus three π§ minus four
equals zero and two π₯ plus two π¦ minus nine π§ plus seven equals zero.

Letβs proceed as we normally would
when solving two simultaneous equations by trying to eliminate one of the
variables. We can eliminate π§ by taking three
times the first equation and adding it to the second equation. This gives five π₯ minus 10π¦ minus
five equals zero. Adding 10π¦ plus five to both sides
and dividing by five gives π₯ purely in terms of π¦: π₯ equals two π¦ plus one. Letβs move this equation up here
for safekeeping. We can now go back to the equations
of the planes and eliminate one of the other variables. We can eliminate π¦ by taking the
first equation and adding two times the second equation. This gives five π₯ minus 15π§ plus
10 equals zero. Adding 15π§ and subtracting 10 from
both sides and then dividing by five gives π₯ purely in terms of π§: π₯ equals three
π§ minus two.

We now have two expressions for π₯,
one in terms of π¦ and one in terms of π§. Since both of these expressions are
equal to π₯, they are also equal to each other. We can therefore rewrite these two
equations as one equation with two equalities: π₯ equals two π¦ plus one equals
three π§ minus two. This is the general equation of the
line of intersection between the two planes.

We cannot reduce the system of
equations any further than this or find values for π₯, π¦, and π§ that uniquely
solve the equations because we have one more unknown than the number of equations
and therefore an infinite number of solutions. However, we are free to set the
value of one variable, which will give us corresponding values of the other two
variables. This will give us one of the
infinitely many solutions to the two equations, which is one point on the line of
intersection.

For example, we can set π§ equal to
one. From this second part of the
equation then, we have two π¦ plus one equals one. Rearranging for π¦ gives π¦ equals
zero. From the other part of the
equation, π₯ equals two π¦ plus one. And since π¦ is equal to zero, π₯
is equal to one. Therefore, π₯ equals one, π¦ equals
zero, and π§ equals one is one of the infinitely many solutions to the two equations
of the two planes. And the point one, zero, one lies
on the line of intersection.

The general equation is not the
only way of describing the line of intersection between two planes. Another way is with a set of
parametric equations, where π₯, π¦, and π§ are each defined separately by an
external parameter. Letβs look at an example of how to
do this.

Find the parametric equations of
the line of intersection between the two planes π₯ plus π§ equals three and two π₯
minus π¦ minus π§ equals negative two.

Recall that a straight line in 3D
space may be described by the set of parametric equations π₯ equals π₯ naught plus
ππ‘, π¦ equals π¦ naught plus ππ‘, and π§ equals π§ naught plus ππ‘. π₯ naught, π¦ naught, π§ naught can
be any point on the line π naught, and π, π, π is a direction vector that is
parallel to the line. These parametric equations are
arbitrary because π naught can be any point on the line that we choose, and π is
just one direction vector that is parallel to the line. Any nonzero scalar multiple of π,
πΌπ, is also parallel to the line.

To find the set of parametric
equations for the line of intersection, we set one of these arbitrary
parameterizations for one of the variables then substitute it into the two equations
of the planes then rearrange the resulting equations to find expressions for the
other two variables, also in terms of the parameter. For our two planes, we have the
equations π₯ plus π§ equals three and two π₯ minus π¦ minus π§ equals negative
two. If we substitute in an arbitrary
parameterization for π₯, we get π₯ naught plus ππ‘ plus π§ equals three and two
times π₯ naught plus ππ‘ minus π¦ minus π§ equals negative two. We can rearrange the first equation
to give π§ purely in terms of the parameter: π§ equals three minus π₯ naught minus
ππ‘.

Remember that π₯ naught and π are
both constants, so π§ is a function of just the parameter π‘. We can now substitute this
expression for π§ into the second equation, which gives two times π₯ naught plus
ππ‘ minus π¦ minus three minus π₯ naught minus ππ‘ equals negative two. Rearranging for π¦ gives π¦ purely
in terms of the parameter: π¦ equals three times π₯ naught plus ππ‘ minus one. We now have π₯, π¦, and π§
expressed purely as functions of the parameter π‘. So we have the arbitrary set of
parametric equations. We are free to choose any values
for π₯ naught and π that we like with the exception that π cannot be equal to zero
because then changing the value of π‘ would not change the position on the line.

If we look at the list of possible
answers, we can see that four of them have π§ equals negative π‘. If we choose this as our parametric
equation for π§, this implies that three minus π₯ naught minus ππ‘ is equal to
negative π‘. So this must mean that π₯ naught is
equal to three and π is equal to one. We can now substitute these values
for π₯ naught and π into the equations for π₯ and π¦. For π₯, we have π₯ equals three
plus π‘, and for π¦ we have π¦ equals three times three plus π‘ minus one, which
simplifies to eight plus three π‘. Our set of parametric equations
therefore matches with answer (c) π₯ equals three plus π‘, π¦ equals eight plus
three π‘, and π§ equals negative π‘.

As we saw in the previous question,
a line in 3D space may be defined by a point on the line and a direction vector
parallel to the line. This hints at another way of
describing a line in 3D space with a vector equation. Letβs look at an example of
this.

Find the vector equation of the
line of intersection between the planes π₯ plus three π¦ plus two π§ minus six
equals zero and two π₯ minus π¦ plus π§ plus two equals zero.

Recall that the vector equation of
a line in 3D space is given by π« equals π« naught plus π‘π. π« naught is the position vector of
a point on the line π₯ naught, π¦ naught, π§ naught. π‘ is a scalar. And π is a direction vector
parallel to the line. This equation is not unique since
we are free to choose any point that we like on the line for π« naught, and any
vector that is a nonzero scalar multiple of π will also be parallel to the
line. So to find the vector equation for
the line of intersection between the two planes, all we need to do is find a point
in both planes with position vector π« naught and a direction vector π that is
parallel to the line of intersection.

Letβs begin by finding a point that
lies in both planes for π« naught. Assuming the line of intersection
is not parallel to any of the coordinate planes, we can choose any value for any one
variable that we like and find corresponding values for the other two variables. This will give us one point that
lies in both planes. Since every one of the possible
answers has π₯ equal to zero in the constant vector, letβs try π₯ equals zero. If the line of intersection happens
to be parallel to the π¦π§-plane, the value of π₯ will be constant and probably not
equal to zero. In which case, setting π₯ equal to
zero will mean that we will not be able to find solutions to the two equations. If this were the case, we would
need to set some other variable equal to some other value.

Fortunately, that is not the case
here. Setting π₯ equal to zero gives us
three π¦ plus two π§ minus six equals zero and negative π¦ plus π§ plus two equals
zero. We can simply rearrange the second
equation to give π¦ equals π§ plus two. And substituting this expression
for π¦ into the first equation gives us three times π§ plus two plus two π§ minus
six equals zero. And rearranging for π§ gives us π§
equals zero. And since π¦ is equal to π§ plus
two, π¦ is equal to two. So by setting π₯ equals zero into
the two equations of the planes, weβve obtained π¦ equals two and π§ equals
zero. So the point zero, two, zero lies
in both planes. We therefore have the position
vector π« naught of a point that lies on the line of intersection zero, two,
zero.

We now need to find a direction
vector π that is parallel to the line of intersection. Since the line of intersection lies
on both planes, its direction vector is parallel to both planes. If we look down the axis of the two
planes intersecting, their line of intersection comes straight out of the
screen. Their normal vectors, however, π§
one and π§ two, both lie in the plane of the screen. The cross product of the two normal
vectors, π§ one cross π§ two, is perpendicular to both π§ one and π§ two and
therefore also either come straight out of the screen or go straight into the
screen. So we can use this as our direction
vector for the line of intersection.

We can obtain the normal vectors π§
one and π§ two by simply reading off the coefficients of the variables in each of
the equations of the planes. π§ one is therefore equal to one,
three, two and π§ two is equal to two, negative one, one. Their cross product, π§ one cross
π§ two, is given by the determinant of the three-by-three matrix π’, π£, π€ followed
by the components of π§ one one, three, two and then the components of π§ two two,
negative one, one, where π’, π£, and π€ are the unit vectors in the π₯-, π¦-, and
π§-directions, respectively. Expanding the determinant along the
top row gives us π’ times three minus negative two minus π£ times one minus four
plus π€ times negative one minus six. Simplifying gives us five π’ plus
three π£ minus seven π€, which as a tuple is five, three, negative seven.

We therefore have our direction
vector for the vector equation of the line of intersection: five, three, negative
seven. Our complete vector equation for
the line of intersection between the two planes is therefore π« equals zero, two,
zero plus π‘ times five, three, negative seven, which matches with answer (d).

Sometimes instead of the line of
intersection between two planes, we may wish to find the point of intersection
between a line and a plane in 3D space. Letβs look at an example of
this.

Find the point of intersection of
the straight line negative three π₯ equals four π¦ minus two equals π§ plus one and
the plane negative three π₯ plus π¦ plus π§ equals 13.

If they are not parallel or
coplanar, a line and a plane in 3D space will intersect at a single point π₯ naught,
π¦ naught, π§ naught. This point is the unique solution
to the equation of the line and the equation of the plane. We have three unknowns π₯, π¦, and
π§. We have one equation, the equation
of the plane. And the equation of the straight
line is effectively two equations since there are two equalities. We can rewrite the equation of the
line as two distinct equations: negative three π₯ equals π§ plus one, and four π¦
minus two equals π§ plus one. We therefore effectively have three
distinct equations for three unknowns. So as long as the line and the
plane are not parallel or coplanar, there should be one unique solution to these
three equations, which is the point where they intersect.

We can rewrite these two equations
to give π₯ and π¦ both purely in terms of π§. π₯ equals negative one-third times
π§ plus one, and π¦ equals one-quarter times π§ plus three. We can then substitute these
expressions for π₯ and π¦ into the equation of the plane, which will give us one
equation for one variable π§, which we can solve to find the value of π§ and
therefore the corresponding values of π₯ and π¦. So substituting in these
expressions into the equation of the plane gives us negative three times negative
one-third times π§ plus one plus one-quarter times π§ plus three plus π§ equals
13.

Distributing the parentheses gives
us π§ plus one plus π§ over four plus three-quarters plus π§ equals 13. Rearranging by collecting terms in
π§ on the left-hand side gives us nine π§ over four equals 45 over four. Multiplying by four and dividing by
nine gives us π§ equals five. We already have expressions for
both π₯ and π¦ in terms of π§, so we can substitute in this value of π§ to give the
values of π₯ and π¦. π₯ is equal to negative one-third
times five plus one, which equals negative two, and π¦ is equal to one-quarter times
five plus three, which is equal to two. The point of intersection between
the line and the plane is therefore negative two, two, five.

Letβs finish this video by
recapping some key points. Two nonparallel planes in 3D space
will intersect over a straight line. This line contains the infinitely
many points π₯, π¦, π§ that satisfy the equations of the two planes. The line of intersection is
one-dimensional. If it is not parallel to any of the
coordinate planes, setting the value of one variable will give corresponding values
for the other two. And finally, a line and a
nonparallel plane in 3D space will intersect at a single point, which is the unique
solution to the equation of the line and the equation of the plane.