### Video Transcript

Find the parametric equations of
the line of intersection between the two planes π₯ plus π§ equals three and two π₯
minus π¦ minus π§ equals negative two.

Recall that a straight line in 3D
space may be described by the set of parametric equations π₯ equals π₯ naught plus
ππ‘, π¦ equals π¦ naught plus ππ‘, and π§ equals π§ naught plus ππ‘. π₯ naught, π¦ naught, π§ naught can
be any point on the line π naught, and π, π, π is a direction vector that is
parallel to the line. These parametric equations are
arbitrary because π naught can be any point on the line that we choose, and π is
just one direction vector that is parallel to the line. Any nonzero scalar multiple of π,
πΌπ, is also parallel to the line.

To find the set of parametric
equations for the line of intersection, we set one of these arbitrary
parameterizations for one of the variables then substitute it into the two equations
of the planes then rearrange the resulting equations to find expressions for the
other two variables, also in terms of the parameter. For our two planes, we have the
equations π₯ plus π§ equals three and two π₯ minus π¦ minus π§ equals negative
two. If we substitute in an arbitrary
parameterization for π₯, we get π₯ naught plus ππ‘ plus π§ equals three and two
times π₯ naught plus ππ‘ minus π¦ minus π§ equals negative two. We can rearrange the first equation
to give π§ purely in terms of the parameter: π§ equals three minus π₯ naught minus
ππ‘.

Remember that π₯ naught and π are
both constants, so π§ is a function of just the parameter π‘. We can now substitute this
expression for π§ into the second equation, which gives two times π₯ naught plus
ππ‘ minus π¦ minus three minus π₯ naught minus ππ‘ equals negative two. Rearranging for π¦ gives π¦ purely
in terms of the parameter: π¦ equals three times π₯ naught plus ππ‘ minus one. We now have π₯, π¦, and π§
expressed purely as functions of the parameter π‘. So we have the arbitrary set of
parametric equations. We are free to choose any values
for π₯ naught and π that we like with the exception that π cannot be equal to zero
because then changing the value of π‘ would not change the position on the line.

If we look at the list of possible
answers, we can see that four of them have π§ equals negative π‘. If we choose this as our parametric
equation for π§, this implies that three minus π₯ naught minus ππ‘ is equal to
negative π‘. So this must mean that π₯ naught is
equal to three and π is equal to one. We can now substitute these values
for π₯ naught and π into the equations for π₯ and π¦. For π₯, we have π₯ equals three
plus π‘, and for π¦ we have π¦ equals three times three plus π‘ minus one, which
simplifies to eight plus three π‘. Our set of parametric equations
therefore matches with answer (c) π₯ equals three plus π‘, π¦ equals eight plus
three π‘, and π§ equals negative π‘.