### Video Transcript

In this video, weβll learn how to
calculate the perpendicular distance between a point and a straight line or between
two parallel lines in space using a formula. And as weβll see, we can use the
same formula in both cases.

As we get started, letβs consider
this first case, finding the perpendicular distance between a point and a line. If this is our point and this is
our line, then that distance would look like this, and we can call it π. To solve for π, there are three
pieces of information weβll need. First, weβll need to know the
coordinates of our point in space. Weβll call that point π. Weβll also need to know the
coordinates of a point on the line. It could be anywhere on the line,
and weβll call that point πΏ. Lastly, weβll need to know the
components of a vector that is parallel to our line, and weβll call this vector
π¬.

In our scenario, weβll let this
point be point π, and letβs say we also know a point on the line. And as we said, this could be
anywhere on the line. And lastly, say we also know the
components of a vector, weβve called it π¬, that is parallel to our line. Our claim is that once we know
these things, we can solve for the minimum distance, the perpendicular distance,
between our point and the line. Hereβs how we can do that. First, weβll set up a vector that
goes from point π to point πΏ. Weβll call this vector ππ. And notice how the vector ππ and
the vector π¬ form two sides of a parallelogram. If we call the area of that
parallelogram π΄, then we can say that the magnitude of the vector ππ crossed with
the vector π¬ is equal to that area.

Now, on our sketch, hereβs
something interesting. If we take the perpendicular
distance π and we slide it so that itβs now in contact with the tail end of the
vector π¬, then that distance π multiplied by the magnitude of the vector π¬ gives
us the area of this rectangle. And the area of this rectangle in
orange and the parallelogram in pink are identical. So not only can we say that the
magnitude of ππ cross π¬ equals π΄, we can also say that π times the magnitude of
π¬ itself equals π΄. That means we can write this
equation here. And if we divide both sides by the
magnitude of the vector π¬, canceling that factor on the left, then we get this
expression for the perpendicular distance π. And as weβve seen, the three things
we need to calculate π are a point in space, a point on our line, and a vector
parallel to the line.

Knowing this, letβs change our
scenario so that now weβre solving for the perpendicular distance between two
parallel lines. The wonderful thing about doing
this is how similar it is to the previous case. Once again, to calculate π, we
need to know three things: a point on line one, weβll call that point πΏ; a point
somewhere on line two, weβll call that point π; and we also need a vector parallel
to both lines. Knowing all this, we can again
define a vector from point π to point πΏ and use it and the vector π¬ in our
equation to solve for π. The best way to learn all this is
through practice. So letβs look now at an
example.

Find, to one decimal place, the
perpendicular distance from point negative three, negative four, zero to the line on
points one, three, one and four, three, two.

Okay, so here we have these two
points in space with coordinates one, three, one and four, three, two. And weβre told that a line lies
along both these points. Along with this, thereβs a third
point whose coordinates we know. And we want to solve for the
perpendicular distance between this third point and the line. Weβll call this distance π.

To begin doing this, we can recall
the equation for the perpendicular distance between a line and a point in space. To carry out this calculation,
weβll need to know the coordinates of a point in space, we have those, as well as
the coordinates of a point on our line, and we actually have two of those. The last thing we do need, though,
to use this equation for π are the components of a vector that is parallel to our
line. Weβre not given that
information.

But notice that if we drew a vector
from one given point on our line to the other, then this vector would be parallel to
our line and we could use it in our equation to solve for π. If we call this vector π¬, we can
see that its components are equal to the coordinates of our first point on the line
β one, three, one β minus those of our second point. What we get is a vector with
components negative three, zero, negative one. Now that we know the vector π¬ that
weβll use in our equation for π, letβs consider this vector ππ. Here, π is a point in
three-dimensional space and πΏ is a point somewhere along our line.

Letβs say that on our line we
choose the point one, three, one to be point πΏ. That means the vector ππ will
look like this. And we can solve for the components
of this vector the same way we did for those of π¬. We subtract the coordinates of
point π from those of point πΏ. And we find that ππ has
components four, seven, and one. We now have all the information
needed to start calculating our perpendicular distance. Weβll begin by computing the cross
product of ππ and π¬. Thatβs equal to the determinant of
this three-by-three matrix. Here, our first row are the π’, π£,
and π€ hat unit vectors. And the second and third rows are
the respective components of vectors ππ and π¬.

Working component by component, we
get π’ hat times seven minus π£ hat times one plus π€ hat times negative 21. We can also write this as the
vector with components seven, negative one, negative 21. These then are the components of
the cross product of ππ and π¬. And now we can take the magnitude
of this cross product and divide it by the magnitude of π¬. The magnitude of ππ cross π¬
equals the square root of seven squared plus negative one squared plus negative 21
squared. And we then divide this by the
magnitude of π¬, which equals the square root of negative three squared plus zero
squared plus negative one squared. Entering this expression in our
calculator and keeping our result to one decimal place, we get an answer of 7.0. Our final answer then is that the
perpendicular distance from the given point to the given line is equal to 7.0 length
units.

Letβs look now at another example
where we find the minimum distance between a point and a line.

Determine, to the nearest
hundredth, the length of the perpendicular drawn from the point negative five,
negative seven, negative 10 to the straight line π₯ plus eight over two equals π¦
minus nine over eight equals π§ plus seven over negative eight.

Okay, so here we have this known
point in space and we also know thereβs a straight line passing through space. And we want to know the length of
the perpendicular from the line to the point. Weβll call this length π. We can recall that π is given by
this expression, where π¬ and ππ are vectors. The condition for vector π¬ is that
it is parallel to our given line. Regarding vector ππ, this is a
vector that points from our given point in space, and weβll call this point π, to a
point that lies somewhere anywhere along our line. And weβll call this point πΏ. This then is what that vector looks
like. We can see that, to solve for the
distance π, weβll need to know the components of a vector parallel to our line, the
coordinates of a point on the line, and the coordinates of a point in space.

Weβre given the coordinates of
point π. And to solve for π¬ and πΏ, weβll
use the equation of our straight line given here. What we want to do is to convert
the form of this equation into whatβs called vector form. Itβs given to us in what is called
symmetric form. And we can write it this way
because all three of these fractions are equal to a scale factor we can call π‘. This means that we can write out
separate equations for the π₯-, π¦-, and π§-coordinates of our line. For example, the fact that π₯ plus
eight over two equals π‘ means that π₯ must be equal to two times π‘ minus
eight. Likewise, since π¦ minus nine over
eight equals π‘, we can say that π¦ equals eight π‘ plus nine. And lastly, π§ plus seven over
negative eight equaling π‘ implies that π§ equals negative eight π‘ minus seven.

Now, our line is expressed in what
is called parametric form. And with just a few small changes,
we can write this in vector form. If we collect the π₯, π¦, and π§
equations together, we can say the π₯-, π¦-, and π§-components of our line are equal
to a vector negative eight, nine, negative seven plus our scale factor π‘ times the
vector two, eight, negative eight. If we collect these π₯-, π¦-, and
π§-components into a single vector weβll call π«, then π« equals the vector negative
eight, nine, negative seven plus the scale factor π‘ times the vector two, eight,
negative eight.

This first vector in our equation
is a vector from the origin of a coordinate frame to a point on the line. That tells us then that the point
with coordinates negative eight, nine, negative seven lies along our line. And then over here, this vector we
multiply by π‘ is a vector that runs along the length of our line. In other words, itβs parallel to
it.

All this means we know the
components of a vector parallel to our line and the coordinates of a point on
it. The vector π¬ has components two,
eight, negative eight, and the point πΏ has coordinates negative eight, nine,
negative seven. Knowing this, we can now define the
components of vector ππ in our scenario. Itβs equal to the coordinates of
point πΏ minus those of point π. We get a vector with components
negative three, 16, three.

Now that we know the components of
π¬ and ππ, weβre ready to calculate the cross product of these vectors. Thatβs equal to the determinant of
this three-by-three matrix. Here, in our first row are the π’,
π£, and π€ hat unit vectors and in the second and third rows the corresponding
components of vectors ππ and π¬, respectively. This is equal to π’ hat times
negative 152 minus π£ hat times 18 plus π€ hat times negative 56 or, in other words,
a vector with components negative 152, negative 18, negative 56.

Weβre now ready to calculate the
magnitude of this cross product and then divide it by the magnitude of π¬. The magnitude of ππ cross π¬
equals the square root of negative 152 squared plus negative 18 squared plus
negative 56 squared. We then divide that by the
magnitude of π¬, itself equal to the square root of two squared plus eight squared
plus negative eight squared. When we evaluate this entire
fraction here, the result, rounded to the nearest hundredth, is 14.19. This then is the length of the
perpendicular from the given point to the given line.

Letβs now look at an example where
we calculate the distance between two parallel lines.

Find, to the nearest hundredth, the
distance between the parallel lines πΏ one, π₯ plus seven over nine equals π¦ plus
one over five equals π§ minus seven over negative six, and πΏ two, π₯ plus three
over nine equals π¦ plus 10 over five equals π§ plus 10 over negative six.

Okay, so here we have these two
lines that are parallel, πΏ one and πΏ two. Our question asks us to solve for
the distance between them, meaning the minimum or perpendicular distance between
these lines. Letβs call that distance π, and we
can recall the mathematical relationship for this distance. What weβll need to know are the
components of a vector, weβve called it π¬, that runs parallel to our two lines. And weβll also need to know the
components of a vector ππ that runs from a point on our second line to a point on
our first line. We can say then that, to calculate
π, the coordinates of points π and πΏ and the components of vector π¬ are what we
need to know.

To solve for this information,
letβs look at the given equations of lines one and two. Starting with line one, this is
given in whatβs called symmetric form. We can write these three fractions
as being equal to one another because we say that theyβre also equal to a scale
factor. We can call it π‘ one. The fact that these three fractions
are all equal to π‘ one means that we can write separate equations for the π₯-, π¦-,
and π§-coordinates of line πΏ one. That is, since π₯ plus seven
divided by nine equals π‘ one, we can write that π₯ equals nine times π‘ one minus
seven. In a similar way, since π¦ plus one
divided by five equals π‘ one, π¦ is equal to five times π‘ one minus one. And then, π§ minus seven over
negative six equaling π‘ one implies that π§ equals negative six π‘ one plus
seven.

Line πΏ one is now written in
whatβs called parametric form. We can combine these three separate
equations into one vector equation. The vector with components π₯, π¦,
π§ is equal to the vector to negative seven, negative one, seven plus our scale
factor π‘ one times the vector nine, five, negative six.

Now, if we take our π₯-, π¦-, and
π§-components and write them as the vector π«, then this equals the vector negative
seven, negative one, seven plus π‘ one times the vector nine, five, negative
six. This first vector is a vector from
the origin of a coordinate system to a point on line πΏ one. That is, the point with coordinates
negative seven, negative one, seven lies on πΏ one. And then, regarding the second
vector, this is a vector that lies along the axis of this line. That means itβs parallel to the
line. And therefore, we can say that this
is our vector π¬.

So far then, we have the components
of a vector parallel to both of our lines. And we also have the coordinates of
a point on our first line. To find the coordinates of a point
on our second line, letβs look at that lineβs equation. Once again, this equation is given
to us in symmetric form. That means we can say that each one
of these fractions is equal to another scale factor weβll call π‘ two. If we once again write out separate
equations for π₯, π¦, and π§, we find that π₯ equals nine π‘ two minus three, π¦
equals five π‘ two minus 10, and π§ equals negative six π‘ two minus 10.

Once again, we can write these
parametric equations as a vector equation. A vector with components π₯, π¦,
and π§ equals the vector negative three, negative 10, negative 10 plus π‘ two times
the vector nine, five, negative six. Written this way, we know that this
vector travels from the origin of a coordinate frame to a point along our line πΏ
two. This tells us that the coordinates
of that point are simply the components of the vector. We can write then that the
coordinates of the point weβve called π are negative three, negative 10, negative
10. Our next step is to use the
coordinates of points πΏ and π to solve for the vector ππ. This is the vector with components
equal to the difference between the coordinates of point πΏ and those of point
π. ππ, therefore, has components
negative four, nine, 17.

We can now move ahead by
calculating the cross product of vectors ππ and π¬. That equals the determinant of this
matrix. And notice that, by column, we have
the π₯-, π¦-, and π§-components of our two vectors. Calculating this cross product, we
get π’ hat times negative 139 minus π£ hat times negative 129 plus π€ hat times
negative 101. We can write this result then as a
vector with components negative 139, positive 129, and negative 101. Now that we know ππ cross π¬,
weβre ready to calculate the magnitude of this cross product and divide it by the
magnitude of π¬.

The magnitude of ππ cross π¬
equals the square root of negative 139 squared plus 129 squared plus negative 101
squared. We divide this by the magnitude of
π¬, which itself is the square root of nine squared plus five squared plus negative
six squared. Calculating this entire fraction
altogether, rounding to the nearest hundredth, we get an answer of 18.03. We say then that the distance
between these two parallel lines is 18.03 length units.

Letβs finish up our lesson now by
summarizing a few key points. In this lesson, we saw that given a
point and a line in space, the perpendicular distance between them is given by this
expression. ππ is a vector going from a point
π in space to a point πΏ on the line in question, and π¬ is a vector thatβs
parallel to that line. We also saw that given two parallel
lines in space, the perpendicular distance between them is given by the same
expression. In this case, π¬ is a vector thatβs
parallel to both lines, π is a point that lies on one line, πΏ is a point that lies
on the other, and, once again, ππ is a vector that goes from point π to point
πΏ.