### Video Transcript

In this video, we will learn how to
calculate the perpendicular distance between a plane and a point, between a plane
and a straight line parallel to it, and between two parallel planes using a
formula.

In order to find the shortest
distance between a point and a plane, we first need to determine exactly what is
meant by the shortest distance between these two geometric objects. Letβs begin by considering the
plane with general equation ππ₯ plus ππ¦ plus ππ§ plus π equals zero, together
with a point with coordinates π₯ sub one, π¦ sub one, π§ sub one.

To find the shortest distance
between these two objects, letβs firstly consider the distance between point π and
a point π
that lies on the plane. We can show that this is not the
shortest distance between point π and the plane by constructing the following right
triangle. We choose a point π on the plane
such that the line segment ππ is perpendicular to the plane. We can then see that the line
segment ππ
is the hypotenuse of the right triangle, which means that it is longer
than the other sides. In particular, it means that the
length of ππ is shorter than ππ
. And as we can construct this
triangle for any point π
that lies on the plane, then the line segment ππ is the
shortest distance between the point π and the plane.

In this video, we will just quote
the formula that can be used to calculate this distance which we will call π·. The shortest, or perpendicular,
distance from some point π with coordinates π₯ sub one, π¦ sub one, π§ sub one and
a plane with equation ππ₯ plus ππ¦ plus ππ§ plus π equals zero is equal to the
absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all
divided by the square root of π squared plus π squared plus π squared.

Whilst it is outside of the scope
of this lesson, this formula can be derived using our knowledge of right angle
trigonometry together with the scalar, or dot, product of two vectors, in this case,
the vectors ππ and ππ. Letβs now consider an example where
we can use this formula to calculate the distance between a point and a plane.

Find the distance between the point
negative five, negative eight, negative six and the plane negative two π₯ plus π¦
plus two π§ equals seven.

In this question, we are asked to
find the distance between a point and a plane. We recall that the distance between
a point and a plane means the perpendicular distance since this is the shortest
distance between the two objects. There is a formula that helps us to
do this. The perpendicular distance π·
between a point π₯ sub one, π¦ sub one, π§ sub one and the plane ππ₯ plus ππ¦ plus
ππ§ plus π equals zero is given by π· is equal to the absolute value of ππ₯ sub
one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π
squared plus π squared plus π squared.

We are given a point with
coordinates negative five, negative eight, negative six. This means that π₯ sub one is equal
to negative five, π¦ sub one equals negative eight, and π§ sub one is equal to
negative six. We note that the equation of the
plane is given in a slightly different format to what is required. By subtracting seven from both
sides of the equation, we have negative two π₯ plus π¦ plus two π§ minus seven
equals zero. Since π, π, and π are the
coefficients of π₯, π¦, and π§, respectively, we have π is equal to negative two,
π is equal to one, and π is equal to two. π is equal to the constant term
negative seven.

We can now substitute these values
into the formula. The numerator simplifies to the
absolute value of 10 plus negative eight plus negative 12 plus negative seven. And the denominator is the square
root of four plus one plus four. And this, in turn, is equal to the
absolute value of negative 17 over root nine. The absolute value of a number is
its distance from zero, so the absolute value of negative 17 is 17. And since the square root of nine
is three, we have π· is equal to 17 over three. And we can therefore conclude that
the distance between the point negative five, negative eight, negative six and the
plane negative two π₯ plus π¦ plus two π§ equals seven is 17 over three length
units.

In this question, we calculated the
distance between a point and a plane. We will now consider how we can
adapt our formula to find the distance between a straight line and a plane. If the equation of our plane is
given in vector form as opposed to general form, we can still use the same formula
to find the shortest distance between a point and a plane. We can use the same process to find
the shortest distance between a straight line and a plane, recalling that if a line
and a plane are not parallel and not coincident, then they intersect, which means
that the distance between them would be zero.

Then, if they are parallel and
distinct, we can show that the shortest distance between them is the perpendicular
distance between any point on the line and the plane. In the diagram shown, we choose an
arbitrary point π with coordinate π₯ sub one, π¦ sub one, π§ sub one that lies on
the line, together with an arbitrary point π
that lies on the plane. Once again, we see that the line
segment ππ
is the hypotenuse of a right triangle, which means that the length of
the line segment ππ is the shortest distance to the plane.

This can be summarized as
follows. The shortest distance π· between a
parallel line and a plane where π₯ sub one, π¦ sub one, π§ sub one is any point on
the line and the plane has the equation the scalar product of vector π« and the
vector π, π, π is equal to negative π is given by capital π· is equal to the
absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all
divided by the square root of π squared plus π squared plus π squared. We will now look at an example
where we need to calculate this distance.

Find the perpendicular distance
between the line π« which is equal to one, two, four plus π‘ multiplied by negative
two, one, four and the plane which is equal to the scalar product of π« and two,
zero, one equals one.

This question involves finding the
perpendicular, or shortest, distance between a line and a plane. Both the line and plane are
currently given in vector form. When considering a line and a
plane, there are two possibilities. Firstly, the line is parallel to
the plane. Or secondly, it intersects the
plane. If the line does intersect the
plane, then the shortest distance between them is equal to zero. This means that the first question
we need to ask is, do the line and plane intersect or are they parallel? Letβs begin by considering the
equation of the line. This can be rewritten as π« is
equal to one minus two π‘, two plus π‘, four plus four π‘.

We can now substitute this vector
into the equation of the plane. We have the dot, or scalar, product
of one minus two π‘, two plus π‘, four plus four π‘ and two, zero, one is equal to
one. Finding the dot product gives us
the following equation, and this simplifies to two minus four π‘ plus four plus four
π‘ equals one. On the left-hand side, the four
π‘βs cancel, and we are left with six equals one. This is incorrect and is therefore
not true for any value of π‘. And we can therefore conclude that
the line and plane do not intersect and must therefore be parallel.

The shortest distance between the
line and the plane can therefore be found by taking any point π that lies on the
line and finding the perpendicular distance to the plane. In order to find the position
vector of any point that lies on the line, we can substitute any value of π‘ into
our equation. For example, when π‘ is equal to
zero, π« is equal to one, two, four. This means that the point with
coordinates one, two, four lies on the line. And we will let this be point
π.

We will now recall the formula that
enables us to calculate the perpendicular, or shortest, distance from a point to a
plane. When the equation of the plane is
written in vector form, as in this case, then the perpendicular distance capital π·
is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one
plus π all divided by the square root of π squared plus π squared plus π
squared. The values of π₯ sub one, π¦ sub
one, and π§ sub one are one, two, and four, respectively. From the vector equation of the
plane, we have π is equal to two, π is equal to zero, and π is equal to one. Since negative π is equal to one,
π is equal to negative one.

Substituting in our values, we have
the distance π· is equal to the absolute value of two multiplied by one plus zero
multiplied by two plus one multiplied by four plus negative one all divided by the
square root of two squared plus zero squared plus one squared. This simplifies to the absolute
value of five divided by the square root of five, which in turn is equal to five
over root five. We can then rationalize the
denominator by multiplying the numerator and denominator by root five. This is equal to five root five
over five, which simplifies to root five. The perpendicular distance between
the given line and plane is root five length units.

Our final example in this video
will involve using the formula to find the distance between two parallel planes. Letβs firstly consider how this can
be done. We will begin by taking an
arbitrary point π on one of the planes. We can then calculate the
perpendicular distance between this point and the other plane as before. In the example that follows, the
equations of the planes will be given in general form. However, it is important to note
that we can use the same formula when the equations are given in vector form.

Find the distance between the two
planes negative π₯ minus two π¦ minus two π§ is equal to negative two and negative
two π₯ minus four π¦ minus four π§ is equal to three.

In this question, we are asked to
find the distance between two planes. This means that we need to find the
perpendicular, or shortest, distance between the two planes. When dealing with two planes, if
they are not parallel, they will intersect and the distance between them will
therefore be equal to zero. This means that the first question
we need to ask is, are the planes parallel? One way to do this is to consider
the normal vectors of the two planes. These are equal to the coefficients
of π₯, π¦, and π§ when the equation of the plane is written in general form. The first plane has normal vector
negative one, negative two, negative two. And the second plane has normal
vector negative two, negative four, negative four. As these two vectors are scalar
multiples of one another, we can conclude that the two planes are parallel.

Letβs now recall how we can find
the distance between two parallel planes. If we pick a point π on one of the
planes, we can calculate the distance π· between the two planes by calculating the
distance between point π and the other plane. This satisfies the formula π· is
equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus
π all divided by the square root of π squared plus π squared plus π squared,
where point π has coordinates π₯ sub one, π¦ sub one, π§ sub one and the plane has
equation ππ₯ plus ππ¦ plus ππ§ plus π equals zero.

We begin by finding the coordinates
of any point that lies on the first plane. One way of doing this is to choose
the point where π₯ equals zero and π¦ equals zero. Substituting these values into the
equation of the first plane, we have negative zero minus two multiplied by zero
minus two π§ is equal to negative two. This simplifies to negative two π§
is equal to negative two. And dividing through by negative
two, we have π§ is equal to one. This means that the coordinates of
one point that lies on the first plane are zero, zero, one. And we now have values of π₯ sub
one, π¦ sub one, and π§ sub one that we can substitute into our formula.

By considering the equation of the
second plane, we see that π is equal to negative two and both π and π are equal
to negative four. These are the coefficients of π₯,
π¦, and π§, respectively. Noting that in order to find π·, we
need the equation of the plane to be equal to zero, we see that π· is equal to
negative three. Substituting our values into the
formula, we have the distance capital π· is equal to the absolute value of negative
two multiplied by zero plus negative four multiplied by zero plus negative four
multiplied by one plus negative three all divided by the square root of negative two
squared plus negative four squared plus negative four squared. This is equal to the absolute value
of negative seven over the square root of 36, which in turn is equal to seven over
six.

We can therefore conclude that the
distance between the two given planes is seven over six length units.

We will now summarize the key
points from this video. The distance π· between the point
π₯ sub one, π¦ sub one, π§ sub one and the plane ππ₯ plus ππ¦ plus ππ§ plus π
equals zero is given by capital π· is equal to the absolute value of ππ₯ sub one
plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π
squared plus π squared plus π squared. We can use the same formula when
the equation of the plane is given in vector form such that the dot, or scalar,
product of vector π« and vector π, π, π is equal to negative π, where π, π, π
is a vector normal to the plane.

We also saw that the distance
between a line parallel to a plane and that plane is equal to the distance between
any point on the line and the plane. In a similar way, the distance
between two parallel planes is equal to the distance between any point on either
plane and the other plane. It is important to note that when
talking about this distance π·, we mean the shortest, or perpendicular, distance
between the two objects.