### Video Transcript

The points π΄ negative eight,
negative nine, negative two; π΅ zero, negative seven, six; and πΆ negative eight,
negative one, negative four form a triangle. Determine in vector form the
equation of the median drawn from πΆ.

In this question, we have three
points π΄, π΅, and πΆ, which are given in three-dimensional space. These three points weβre told form
a triangle. Weβre told that there is a median
drawn from πΆ, 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 π΄π΅πΆ, the median from πΆ 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 π΄π΅. Letβs define this with the letter
π. 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 π of π₯ one,
π¦ one, π§ one and π₯ two, π¦ two, π§ two, we have that π is equal to π₯ one plus
π₯ two over two, π¦ one plus π¦ two over two, π§ one plus π§ two over two. When we fill our values into this
formula, we need to make sure weβre using the values for π΄ and π΅ as, after all, we
need to find the midpoint of π΄π΅.

Note that when weβre plugging in
our values, it doesnβt matter which point we use with our π₯ one, π¦ one, π§ one
values or the π₯ two, π¦ two, π§ two values. So we have that the midpoint π 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 π
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 π« equals π« sub zero plus π‘π―, where π« is the position
vector of a general point on the line, π« sub zero is the position vector of a given
point on the line, and π― is the direction vector. π‘ is a scalar multiple.

Letβs think about what would happen
if we model these three points in three-dimensional space. Weβd have the triangle π΄π΅πΆ and
the median, which would be the line segment of πΆπ. So when it comes to writing the
median in vector form, the position vector can be the point πΆ. But we still need to work out the
direction vector of ππ. To find the vector ππ, we
subtract the starting point πΆ from the terminal point π. So we have negative four subtract
negative eight, negative eight subtract negative one, and two subtract negative
four. Simplifying this, we have that
vector ππ 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. π« sub zero will be the position
vector representing point πΆ. Vector π― will be represented by
the vector ππ. Therefore, the answer for the
equation of the median from πΆ is π« equals negative eight, negative one, negative
four plus π‘ four, negative seven, six.