### Video Transcript

In this video, weβll learn how to
find the equation of a straight line that is parallel or perpendicular to another
one in space and find the intersection point between two lines.

We might recall that we can
characterize a line in three dimensions in vector form, using two pieces of
information. We need a point on the line and a
nonzero vector thatβs parallel to that line. Now this is called the direction
vector. And vectors are parallel when
theyβre scalar multiples of one another. Specifically the vector equation of
a line in three dimensions is given by π« of π‘ equals π₯ nought, π¦ nought, π§
nought plus π‘ times π, π, π. Where π₯ nought, π¦ nought, π§
nought is the position vector of the point that belongs to the line and π, π, π
is the nonzero vector that itβs parallel to.

Now naturally, the components of
the direction vector and the coordinates of the point that the line passes through
explicitly appear in the equation of the line in a variety of forms. So letβs recall the other two. The Cartesian form is π₯ minus π₯
nought over π equals π¦ minus π¦ nought over π equals π§ minus π§ nought over π
for nonzero values of π, π, and π. And the parametric form is π₯
equals π₯ nought plus ππ‘, π¦ equals π¦ nought plus ππ‘, and π§ equals π§ nought
plus ππ‘.

For instance, suppose we know that
a line passes through the point negative one, negative five, four and itβs parallel
to the vector negative three, five, one. We can define π₯ nought to be equal
to negative one, π¦ nought to be equal to negative five, and π§ nought to be equal
to four. And of course, itβs parallel to
this nonzero vector. So π is negative three, π is
five, and π is one. This means that the vector equation
of the line is π« of π‘ equals negative one, negative five, four plus π‘ times
negative three, five, one.

So, given all of this information,
how do we determine whether two lines given in vector form are parallel? Well, it follows that it doesnβt
really matter the point through which the lines pass, but if their direction vectors
are parallel, if theyβre traveling in the same direction, then the lines themselves
must also be parallel. And of course, whilst weβve
represented this in two dimensions, this statement, of course, holds for vectors in
three and in fact higher dimensions. We say that two lines are parallel
then if their direction vectors are parallel. So weβll begin by demonstrating how
to find the vector form of the equation of a line when we know itβs parallel to a
given line.

Find the vector form of the
equation of the straight line passing through the point π΄ two, five, five and
parallel to the straight line passing through the two points π΅ negative three,
negative two, negative six and πΆ five, zero, negative nine.

Weβre told to give the vector form
of the equation of the line. So letβs begin by reminding
ourselves what that looks like. Given a straight line which passes
through the point π₯ nought, π¦ nought, π§ nought and which is parallel to the
vector π, π, π, its equation is π« equals the vector π₯ nought, π¦ nought, π§
nought plus π‘ times the vector π, π, π. And we call this vector π, π, π
the position vector of the line. We quite clearly have the point
through which our line passes through. Thatβs π΄ two, five, five. So according to the definition of
the vector equation of a line, we can define π₯ nought to be equal to two, π¦ nought
to be equal to five, and π§ nought to be equal to five.

But how do we find the direction
vector of our line? We know itβs parallel to the line
that passes through points π΅ and πΆ. So letβs find the vector π©π. And of course, the vector π©π is
given by the vector ππ minus the vector ππ©. Since the coordinates themselves
tell us the position of each point with respect to the origin, then the vector ππ
is simply the vector five, zero, negative nine and the vector ππ© is negative
three, negative two, negative six. And so, we find the difference
between these two vectors by simply subtracting their individual components. That means to find the first
component we subtract negative three from five and we get eight.

Similarly, our second component is
zero minus negative two, which is two. Then our third component is
negative nine minus negative six, which is negative three. And so the vector π©π is the
vector eight, two, negative three. Now, of course, if two vectors are
parallel, that tells us that one vector is a scalar multiple of the other. And so the direction vector of our
line must be some scalar multiple of the vector eight, two, negative three. Now, of course, that scalar
multiple can be one. So weβre going to define the
direction vector of our line, letβs call that π, to be equal to eight, two,
negative three.

And so, of course, all that remains
is to substitute what we know about our line into the general formula for the vector
equation of a straight line. And so the vector equation is π«
equals two, five, five plus π‘ times eight, two, negative three.

In our first example, we
demonstrated how to work with parallel lines. But if two lines in two dimensions
are not parallel nor are they coincident β in other words, lie on top of one another
β it follows that they must intersect at a point. This is not, however, the case in
three dimensions. In this case, they can actually
also be skewed. That is, they donβt intersect, nor
are they parallel to one another. Now of course, if weβre given that
the lines intersect, then we should be able to find the point of intersection of two
lines given to us in three dimensions. Letβs see what that looks like.

Find the Cartesian form of the
equation of the straight line passing through the origin and the intersection point
of the two straight lines πΏ one, which is π« equals one, one, negative two plus π‘
one, four, three, and πΏ two, which is π₯ equals three and π¦ minus five over
negative four equals π§ minus three over negative one.

Remember, if we know that a
straight line passes through the point π₯ nought, π¦ nought, π§ nought and itβs
parallel to the vector π, π, π. Then its Cartesian form is π₯ minus
π₯ nought over π equals π¦ minus π¦ nought over π equals π§ minus π§ nought over
π. So it should be quite clear that to
find the equation of our line, weβre going to need to find a point through which it
passes and a vector which is parallel to it. In fact, weβre told it passes
through the origin, which has coordinates zero, zero, zero. So we can define π₯ nought to be
equal to zero, π¦ nought to be equal to zero, and π§ nought to be equal to zero.

Now since we know that our vector
passes through the origin and the intersection point of our lines πΏ sub one and πΏ
sub two, it follows that weβre going to need to find the value of that intersection
point. Once we have that, we can use
information about that point and the origin to find its direction vector.

Letβs begin by writing each
equation in parametric form, since thatβs generally easier to work with. To achieve this, we distribute the
π‘ over the second vector and then we add the individual components. So π₯ is one plus one times π‘. Thatβs one plus π‘. π¦ is one plus four π‘. And π§ is given by negative two
plus three π‘. For the second line, weβre already
told that π₯ is equal to three, but weβre going to need to do some work with the
second half of this equation. Weβll set each part equal to π‘
since this is equivalent. And then we will rearrange to make
π¦ and π§ the subject, respectively.

Multiplying by negative four and
then adding five to both sides of our first equation, we find π¦ is five minus four
π‘. Similarly, we find that π§ is equal
to three minus π‘. And so we have both of our
equations represented in parametric form. So we know that our lines intersect
at some point π₯, π¦, π§. So this point must be given by some
parameter value π‘ β letβs let that be π‘ one for our first line β and some other
parameter value; letβs let that be π‘ sub two for our second line. Then weβll substitute each value
into the parametric form of the equations of our line and then set them equal to one
another.

For our π₯-coordinate, thatβs one
plus π‘ sub one is simply equal to three, which we can solve to find that π‘ sub one
is equal to two. Then, for our π¦-coordinate, one
plus four π‘ sub one is equal to five minus four π‘ sub two. We then substitute π‘ sub one
equals two into this equation. And we get one plus eight equals
five minus four π‘ sub two. That simplifies to four π‘ sub two
equals negative four. So π‘ sub two must be equal to
negative one. And it should become a quite
apparent that we donβt actually need to perform any further calculations here,
although we could substitute this into the equivalent equation for π§ just to check
that it works.

If we substitute π‘ sub one equals
two and π‘ sub two equals negative one, we get four equals four, which indicates to
us that weβve probably done our calculations correctly. And so we choose one of these
parameters, and we substitute it into the respective equation to find the point of
intersection. Letβs use π‘ sub one equals two and
substitute it into the formulae for πΏ sub one. We get one plus two for the
π₯-coordinate, one plus four times two to the π¦-coordinate, and negative two plus
three times two for our π§-coordinate. And so the point of intersection is
three, nine, four.

Now, if we let this point of
intersection be equal to point π, then we know that the direction vector, our line,
is the vector that joins π to π, the origin, to point π. And thatβs simply three, nine,
four. And so, if we go back right to the
beginning and recall that formula that we had for the Cartesian equation of the
line, we can define π to be equal to three, π to be equal to nine, and π to be
equal to four. Letβs substitute all of that into
the equation of a straight line. And we find that the equation is π₯
over three equals π¦ over nine, which equals π§ over four.

So far, weβve covered the cases of
lines being parallel, not parallel, coincident, or skew. If two lines intersect, however,
there is another final possibility. That is, they might be
perpendicular or orthogonal to one another. In other words, they meet at an
angle of 90 degrees. Letβs define this formally and say
that two lines are perpendicular if they intersect at a point and their direction
vectors are perpendicular. Here, itβs also useful to remind
ourselves that two vectors π and π are perpendicular if and only if their scalar
or dot product is equal to zero. So here, π dot π equals zero. In our next example, weβll
demonstrate an application of these definitions.

Consider the two lines π₯ equals
four plus two π‘, π¦ equals six plus π‘, π§ equals two minus two π‘, and π« equals
six, seven, zero plus π‘ times five, four, seven. Determine whether they are parallel
or perpendicular.

Now we can determine whether two
lines are parallel or perpendicular by considering their direction vectors. And so what weβre going to do is
begin by rewriting the equation of our first line in vector form so we can then
extract its direction vector. And of course, thatβs of the form
π₯ nought, π¦ nought, π§ nought plus π‘ times π, π, π, where π, π, π is the
direction vector and π₯ nought, π¦ nought, π§ nought is the point which the line
passes through. To achieve this, we begin by
writing each expression as the component of our vector. Then we separate the components as
shown. So we get four, six, two plus two
π‘, π‘, negative two π‘. And then we take out that constant
factor of π‘. So the vector form of our first
line is four, six, two plus π‘ times two, one, negative two.

So letβs write down then the
direction vector of each of our lines. Letβs call our first line πΏ sub
one. Its direction vector is two, one,
negative two. Then the direction vector of our
second line, weβll call that πΏ sub two, is five, four, seven. And then we recall that if two
vectors are parallel, specifically direction vectors, then one is a scalar multiple
of the other. So for our lines to be parallel, we
can say that the direction vector two, one, negative two can be written as some
scalar multiple π of the vector five, four, seven.

Now, if we look at the second
component one and four, we might deduce that for this to be true, π would have to
be equal to one-quarter, since one-quarter of four is equal to one. But one-quarter of five is not
equal to two, and a quarter of seven is not equal to negative two. And so this is not true. These vectors cannot be written as
scalar multiples of one another, and so the lines cannot be parallel. We might deduce then that theyβre
perpendicular. But weβre going to check. For two vectors to be
perpendicular, of course, their dot product or scalar products have to be equal to
zero. They also have to intersect at a
point, but weβll deal with that in a moment.

Letβs just double-check that their
dot product is indeed equal to zero. Well, their dot product or their
scalar product is two times five plus one times four plus negative two times seven,
which is indeed equal to zero. So this element so far is good. We do, of course, need to
double-check that they intersect at a point. Now if the two lines intersect at a
point, there must be some parameter values π‘ sub one in our first equation and π‘
sub two in our second that make the π₯-coordinates, π¦-coordinates, and π§
coordinates, respectively, equal.

For our π₯-coordinates, letβs say
thatβs four plus two π‘ sub one is equal to six plus five π‘ sub two. For our π¦-coordinate, itβs six
plus π‘ sub one equals seven plus four π‘ sub two. And then we have a corresponding
equation for our π§-coordinate. Solving any two of these equations
simultaneously, we obtain π‘ sub one equals one and π‘ sub two equals zero. Of course, we can check these by
then substituting into the third equation, the third one that we hadnβt used. Substituting π‘ sub one equals one
into the equation of our first line and we find that this point of intersection is
four plus two, six plus one, two minus two, which is six, seven, zero.

Letβs double-check. We get the same values when we
substitute into πΏ sub two. As expected, we do indeed get the
value of six, seven, zero. And of course, if we had
substituted π‘ sub one equals one and π‘ sub two equal zero into that third equation
that we didnβt use, we would have expected this result. So since the direction vectors of
the two lines are perpendicular and they intersect at a point, we can say that the
lines themselves must also be perpendicular.

Letβs recap some of the key
concepts from this lesson. In this video, we learned that two
lines are parallel if their direction vectors are parallel. We also saw that theyβre
perpendicular if they intersect at a point and their direction vectors themselves
are perpendicular. If two parallel lines intersect at
a point, then they must overlap entirely. And in this case, we call them
coincident. And in three dimensions, we have
skew lines. Now these are not parallel, but
they also donβt intersect at any point.