### Video Transcript

In this video, we’ll see how we can
find the equation of a straight line in space. So that means we’re looking at
coordinates in three dimensions rather than just in two dimensions. We’ll see how we can write this
equation in Cartesian form, which is sometimes called the general form. And we’ll also see how we can write
it in vector form. Let’s begin by having a look at the
vector form.

The vector form of a line can be
described as 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫, 𝐫 sub zero, and 𝐯 are all
vectors. 𝐫 is the position vector of any
general point on the line. 𝐫 sub zero is the position vector
of a given point on the line. 𝐯 is the direction vector of or
along the line. And 𝑡 is a scalar multiple. Vector form can be used in both two
dimensions and three dimensions. The difference is that in three
dimensions, all of our vectors will have 𝑥-, 𝑦-, and 𝑧-components. When we’re describing a line in
vector form, remember that we’re thinking about how we navigate from the origin to a
specific point. And then we’re moving along that
line in scalar multiples of the vector 𝐯.

We’ll now have a look at a question
where we need to write the vector equation of a line given a point on the line and a
direction vector.

Give the vector equation of the
line through the point three, seven, negative seven with direction vector zero,
negative five, seven.

We should remember that when we
need to write an equation in vector form, it will be in the form 𝐫 equals 𝐫 sub
zero plus 𝑡𝐯, where 𝐫 is the position vector of a general point on the line, 𝐫
sub zero is a position vector of a given point on the line, and 𝐯 is the direction
vector. 𝑡 is a scalar multiple. If we look at the information that
we’re given in the question, we can see that we have a direction vector. And we’ve got a point on the line
which can be written as a position vector. As we navigate from the origin to
the point three, seven, negative seven, then we can write this as the position
vector three, seven, negative seven.

We can then simply plug in these
two vectors into the vector form. 𝐫 equals the position vector
three, seven, negative seven plus 𝑡 times the direction vector zero, negative five,
seven. And so that’s the answer for the
vector equation of the line.

In the following question, we’ll
see how we can calculate a direction vector given two points.

Find the direction vector of the
straight line passing through 𝐴 one, negative two, seven and 𝐵 four, negative one,
three.

In this question, we’re given the
position vectors of two points in space, 𝐴 and 𝐵, and we’re asked to find the
direction vector. When we want to find a direction
vector 𝐀𝐁, 𝐴 is the starting point and 𝐵 is the terminal point, we subtract the
starting point from the terminal point. In order to find the direction
vector, we can subtract each of the 𝑥-, 𝑦-, and 𝑧-components in 𝐴 from those in
𝐵. To begin then, we’ll have four
subtract one, giving us three. Then we’ll have negative one
subtract negative two, which is equivalent to negative one plus two, which is
one. And finally, we’ll have three
subtract seven, giving us negative four. And so we’ve got the answer for the
direction vector of 𝐝 as three, one, negative four.

In this question, however, we
didn’t necessarily need to find the direction vector 𝐀𝐁. We could also have found the
direction vector 𝐁𝐀. In this case, we would’ve got the
vector inverse of 𝐝 equals negative three, negative one, four, which would also
have been a valid answer.

In the next question, we’ll see a
slightly more complex example where we need to find the vector equation of a median
of a triangle drawn in three-dimensional space.

The points 𝐴 negative eight,
negative nine, negative two; 𝐵 zero, negative seven, six; and 𝐶 negative eight,
negative one, negative four form a triangle. Determine in vector form the
equation of the median drawn from 𝐶.

In this question, we have three
points 𝐴, 𝐵, and 𝐶, which are given in three-dimensional space. These three points we’re told form
a triangle. We’re told that there is a median
drawn from 𝐶, and so it would be useful to recall that a median is a line segment
joining a vertex to the midpoint of the opposite side. For example, if we drew this
two-dimensional triangle 𝐴𝐵𝐶, the median from 𝐶 would look like this.

Perhaps the best way to begin this
question is to see if we can find the point that is the midpoint of 𝐴𝐵. Let’s define this with the letter
𝑀. The formula to find the midpoint of
two points in space is very similar to that which we might use for two coordinates
in two-dimensional space. To find the midpoint 𝑀 of 𝑥 one,
𝑦 one, 𝑧 one and 𝑥 two, 𝑦 two, 𝑧 two, we have that 𝑀 is equal to 𝑥 one plus
𝑥 two over two, 𝑦 one plus 𝑦 two over two, 𝑧 one plus 𝑧 two over two. When we fill our values into this
formula, we need to make sure we’re using the values for 𝐴 and 𝐵 as, after all, we
need to find the midpoint of 𝐴𝐵.

Note that when we’re plugging in
our values, it doesn’t matter which point we use with our 𝑥 one, 𝑦 one, 𝑧 one
values or the 𝑥 two, 𝑦 two, 𝑧 two values. So we have that the midpoint 𝑀 is
equal to negative eight plus zero over two, negative nine plus negative seven over
two, and negative two plus six over two. Simplifying this, we have that 𝑀
is equal to negative four, negative eight, two. We can now clear some space so we
can begin to think about the vector form of the equation of this median. The vector form of an equation can
be written in the form 𝐫 equals 𝐫 sub zero plus 𝑡𝐯, where 𝐫 is the position
vector of a general point on the line, 𝐫 sub zero is the position vector of a given
point on the line, and 𝐯 is the direction vector. 𝑡 is a scalar multiple.

Let’s think about what would happen
if we model these three points in three-dimensional space. We’d have the triangle 𝐴𝐵𝐶 and
the median, which would be the line segment of 𝐶𝑀. So when it comes to writing the
median in vector form, the position vector can be the point 𝐶. But we still need to work out the
direction vector of 𝐂𝐌. To find the vector 𝐂𝐌, we
subtract the starting point 𝐶 from the terminal point 𝑀. So we have negative four subtract
negative eight, negative eight subtract negative one, and two subtract negative
four. Simplifying this, we have that
vector 𝐂𝐌 is equal to four, negative seven, six.

Now we have all the information
that we need to plug in to the vector form of the line. 𝐫 sub zero will be the position
vector representing point 𝐶. Vector 𝐯 will be represented by
the vector 𝐂𝐌. Therefore, the answer for the
equation of the median from 𝐶 is 𝐫 equals negative eight, negative one, negative
four plus 𝑡 four, negative seven, six.

So far in this video, we’ve looked
at equations of lines in vector form. Now we’ll think about how we change
a line given in vector form to one in Cartesian form. You may be confused by the
terminology of Cartesian form, but a line in two dimensions in Cartesian form can be
written in the form 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope or gradient and 𝑏
is the 𝑦-intercept. But of course, it’s different in
three-dimensional space as we need an equation that describes the 𝑥-, 𝑦-, and
𝑧-variables.

So, to find the equation of a line
in Cartesian form, we can say that given the equation of a line with direction
vector 𝐯 equals 𝑙, 𝑚, 𝑛 which passes through the point 𝑥 sub one, 𝑦 sub one,
𝑧 sub one, then it’s given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub
one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero
real numbers. So we can see how this information
of an equation with a point and a direction, i.e., one in vector form, can be
changed into one in Cartesian form. We’ll now have a look at two
questions where we could apply this formula.

Give the Cartesian equation of the
line 𝐫 equals negative three, negative two, negative two plus 𝑡 four, two,
four.

In this question, we’re given this
equation in vector form. Negative three, negative two,
negative two is the position vector of a given point, and four, two, four is the
direction vector. In order to change the equation in
vector form into an equation in Cartesian form, there is a formula we can apply. The equation of a line with
direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through 𝑥 sub one, 𝑦 sub one, 𝑧
sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦 sub one over 𝑚
equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero real
numbers.

We now need to take the direction
vector four, two, four to have the values of 𝑙, 𝑚, and 𝑛, respectively. We can do the same and designate
the coordinate 𝑥 sub one, 𝑦 sub one, 𝑧 sub one with the values negative three,
negative two, negative two. Plugging these values into the
formula, we have 𝑥 minus negative three over four equals 𝑦 minus negative two over
two equals 𝑧 minus negative two over four. Simplifying the numerators, we have
𝑥 plus three over four equals 𝑦 plus two over two equals 𝑧 plus two over
four. And that’s the answer for the
Cartesian equation of the given line.

Let’s have a look at one final
question.

Find the Cartesian form of the
equation of the straight line passing through the points negative seven, negative
three, negative seven and negative three, negative 10, negative four.

In this question, although we’re
asked for the Cartesian form of the equation, it might be useful to think about what
this line would be like in a vector form. If we considered our two points as
𝐴 and 𝐵 and we wanted to find the direction vector of 𝐀𝐁, then we would subtract
all the points on our starting point 𝐴 from those on 𝐵. So we’d have 𝐀𝐁 is equal to
negative three subtract negative seven, which is equivalent to negative three plus
seven, giving us four. Negative 10 subtract negative three
is equivalent to negative 10 plus three, which is negative seven. And then we’d have negative four
subtract negative seven, which gives us three.

Now that we have a direction vector
and a point on a line, we can find the Cartesian form of the equation of the line
joining these two points. We should remember that the
equation of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through
the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙
equals 𝑦 minus 𝑦 sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛. Note that 𝑙, 𝑚, and 𝑛 are
nonzero real numbers.

We can now use the direction vector
of 𝐀𝐁 for the values of 𝑙, 𝑚, and 𝑛 and the point negative seven, negative
three, negative seven for the values 𝑥 sub one, 𝑦 sub one, and 𝑧 sub one. Filling these values into the
formula, we have 𝑥 subtract negative seven over four equals 𝑦 subtract negative
three over negative seven equals 𝑧 subtract negative seven over three. Simplifying the numerators gives us
the answer in Cartesian form 𝑥 plus seven over four equals 𝑦 plus three over
negative seven equals 𝑧 plus seven over three.

We can now summarize the key points
of this video. Firstly, we saw the equation of a
line given in vector form as 𝐫 equals 𝐫 sub zero plus 𝑡𝐯. 𝐫 is the position vector of a
general point on the line, 𝐫 sub zero is the position vector of a given point on
the line, and 𝐯 is the direction vector. 𝑡 is a scalar multiple. To find a direction vector 𝐀𝐁, we
subtract the starting point 𝐴 from the terminal point 𝐵.

Finally, we saw that the equation
of a line with direction vector 𝐯 equals 𝑙, 𝑚, 𝑛 that passes through 𝑥 sub one,
𝑦 sub one, 𝑧 sub one is given by 𝑥 minus 𝑥 sub one over 𝑙 equals 𝑦 minus 𝑦
sub one over 𝑚 equals 𝑧 minus 𝑧 sub one over 𝑛, where 𝑙, 𝑚, and 𝑛 are nonzero
real numbers. This final formula is very useful
for changing the equation of a line given in vector form to an equation given in
Cartesian form.