### Video Transcript

A cube of a side length of three
sits with a vertex at the origin and three sides along the positive axes. Find the parametric equations of
the main diagonal from the origin.

Letβs begin by making a sketch of
the cube. The main diagonal of the cube goes
from the vertex at the origin, which has coordinates zero, zero, zero, to the vertex
furthest from this, that is, the vertex at the point with coordinates three, three,
three. Now recalling that the parametric
equations of a line in space are π₯ is equal to π₯ sub zero plus ππ‘, π¦ is equal
to π¦ sub zero plus ππ‘, and π§ is equal to π§ sub zero plus ππ‘. And thatβs where π₯ sub zero, π¦
sub zero, π§ sub zero is a point on the line, the vector π, π, π is a direction
vector, and π‘ is a real parameter that varies between negative and positive β.

Now, if we take the origin, thatβs
the point zero, zero, zero, to be the point on the line, thatβs π₯ sub zero, π¦ sub
zero, π§ sub zero, then our parametric equations become π₯ is equal to zero plus
ππ‘, π¦ is zero plus ππ‘, and π§ is zero plus ππ‘. That is, π₯ is ππ‘, π¦ is ππ‘,
and π§ is ππ‘.

So now, we need to find the
coordinates of a direction vector, that is, the components of our direction vector
π, π, and π. And if we have two points on a
line, thatβs π₯ sub one, π¦ sub one, π§ sub one and π₯ sub two, π¦ sub two, π§ sub
two, then these components are given by π is equal to π₯ sub two minus π₯ sub one,
π is equal to π¦ sub two minus π¦ sub one, and π is π§ sub two minus π§ sub
one. So now, if this time we take the
origin as our point π₯ sub one, π¦ sub one, π§ sub one and the vertex at the end of
the main diagonal as the point π₯ sub two, π¦ sub two, π§ sub two β so that has
coordinates three, three, three β then we have π is equal to three minus zero, π
is three minus zero, and π is three minus zero. And so our direction vector has
components three, three, three. And substituting these values into
our parametric equations, we have π₯ is three π‘, π¦ is three π‘, and π§ is three
π‘. And these are the parametric
equations of the main diagonal of the cube.

Itβs worth noting that we couldβve
used the point three, three, three as the point on the line. And for the direction vector, we
couldβve chosen any vector parallel to the line. We could also have restricted the
values of the parameter π‘. In fact, the line we found extends
across the whole of space, whereas if weβd restricted the parameter π‘ to take
values between zero and one only, our equations would describe only the main
diagonal of the cube.