### Video Transcript

Find the vector equation of the
line of intersection between the planes π₯ plus three π¦ plus two π§ minus six
equals zero and two π₯ minus π¦ plus π§ plus two equals zero.

Recall that the vector equation of
a line in 3D space is given by π« equals π« naught plus π‘π. π« naught is the position vector of
a point on the line π₯ naught, π¦ naught, π§ naught. π‘ is a scalar. And π is a direction vector
parallel to the line. This equation is not unique since
we are free to choose any point that we like on the line for π« naught, and any
vector that is a nonzero scalar multiple of π will also be parallel to the
line. So to find the vector equation for
the line of intersection between the two planes, all we need to do is find a point
in both planes with position vector π« naught and a direction vector π that is
parallel to the line of intersection.

Letβs begin by finding a point that
lies in both planes for π« naught. Assuming the line of intersection
is not parallel to any of the coordinate planes, we can choose any value for any one
variable that we like and find corresponding values for the other two variables. This will give us one point that
lies in both planes. Since every one of the possible
answers has π₯ equal to zero in the constant vector, letβs try π₯ equals zero. If the line of intersection happens
to be parallel to the π¦π§-plane, the value of π₯ will be constant and probably not
equal to zero. In which case, setting π₯ equal to
zero will mean that we will not be able to find solutions to the two equations. If this were the case, we would
need to set some other variable equal to some other value.

Fortunately, that is not the case
here. Setting π₯ equal to zero gives us
three π¦ plus two π§ minus six equals zero and negative π¦ plus π§ plus two equals
zero. We can simply rearrange the second
equation to give π¦ equals π§ plus two. And substituting this expression
for π¦ into the first equation gives us three times π§ plus two plus two π§ minus
six equals zero. And rearranging for π§ gives us π§
equals zero. And since π¦ is equal to π§ plus
two, π¦ is equal to two. So by setting π₯ equals zero into
the two equations of the planes, weβve obtained π¦ equals two and π§ equals
zero. So the point zero, two, zero lies
in both planes. We therefore have the position
vector π« naught of a point that lies on the line of intersection zero, two,
zero.

We now need to find a direction
vector π that is parallel to the line of intersection. Since the line of intersection lies
on both planes, its direction vector is parallel to both planes. If we look down the axis of the two
planes intersecting, their line of intersection comes straight out of the
screen. Their normal vectors, however, π§
one and π§ two, both lie in the plane of the screen. The cross product of the two normal
vectors, π§ one cross π§ two, is perpendicular to both π§ one and π§ two and
therefore also either come straight out of the screen or go straight into the
screen. So we can use this as our direction
vector for the line of intersection.

We can obtain the normal vectors π§
one and π§ two by simply reading off the coefficients of the variables in each of
the equations of the planes. π§ one is therefore equal to one,
three, two and π§ two is equal to two, negative one, one. Their cross product, π§ one cross
π§ two, is given by the determinant of the three-by-three matrix π’, π£, π€ followed
by the components of π§ one one, three, two and then the components of π§ two two,
negative one, one, where π’, π£, and π€ are the unit vectors in the π₯-, π¦-, and
π§-directions, respectively. Expanding the determinant along the
top row gives us π’ times three minus negative two minus π£ times one minus four
plus π€ times negative one minus six. Simplifying gives us five π’ plus
three π£ minus seven π€, which as a tuple is five, three, negative seven.

We therefore have our direction
vector for the vector equation of the line of intersection: five, three, negative
seven. Our complete vector equation for
the line of intersection between the two planes is therefore π« equals zero, two,
zero plus π‘ times five, three, negative seven, which matches with answer (d).