### Video Transcript

Consider the two lines π₯ equals
four plus two π‘, π¦ equals six plus π‘, π§ equals two minus two π‘, and π« equals
six, seven, zero plus π‘ times five, four, seven. Determine whether they are parallel
or perpendicular.

Now we can determine whether two
lines are parallel or perpendicular by considering their direction vectors. And so what weβre going to do is
begin by rewriting the equation of our first line in vector form so we can then
extract its direction vector. And of course, thatβs of the form
π₯ nought, π¦ nought, π§ nought plus π‘ times π, π, π, where π, π, π is the
direction vector and π₯ nought, π¦ nought, π§ nought is the point which the line
passes through. To achieve this, we begin by
writing each expression as the component of our vector. Then we separate the components as
shown. So we get four, six, two plus two
π‘, π‘, negative two π‘. And then we take out that constant
factor of π‘. So the vector form of our first
line is four, six, two plus π‘ times two, one, negative two.

So letβs write down then the
direction vector of each of our lines. Letβs call our first line πΏ sub
one. Its direction vector is two, one,
negative two. Then the direction vector of our
second line, weβll call that πΏ sub two, is five, four, seven. And then we recall that if two
vectors are parallel, specifically direction vectors, then one is a scalar multiple
of the other. So for our lines to be parallel, we
can say that the direction vector two, one, negative two can be written as some
scalar multiple π of the vector five, four, seven.

Now, if we look at the second
component one and four, we might deduce that for this to be true, π would have to
be equal to one-quarter, since one-quarter of four is equal to one. But one-quarter of five is not
equal to two, and a quarter of seven is not equal to negative two. And so this is not true. These vectors cannot be written as
scalar multiples of one another, and so the lines cannot be parallel. We might deduce then that theyβre
perpendicular. But weβre going to check. For two vectors to be
perpendicular, of course, their dot product or scalar products have to be equal to
zero. They also have to intersect at a
point, but weβll deal with that in a moment.

Letβs just double-check that their
dot product is indeed equal to zero. Well, their dot product or their
scalar product is two times five plus one times four plus negative two times seven,
which is indeed equal to zero. So this element so far is good. We do, of course, need to
double-check that they intersect at a point. Now if the two lines intersect at a
point, there must be some parameter values π‘ sub one in our first equation and π‘
sub two in our second that make the π₯-coordinates, π¦-coordinates, and π§
coordinates, respectively, equal.

For our π₯-coordinates, letβs say
thatβs four plus two π‘ sub one is equal to six plus five π‘ sub two. For our π¦-coordinate, itβs six
plus π‘ sub one equals seven plus four π‘ sub two. And then we have a corresponding
equation for our π§-coordinate. Solving any two of these equations
simultaneously, we obtain π‘ sub one equals one and π‘ sub two equals zero. Of course, we can check these by
then substituting into the third equation, the third one that we hadnβt used. Substituting π‘ sub one equals one
into the equation of our first line and we find that this point of intersection is
four plus two, six plus one, two minus two, which is six, seven, zero.

Letβs double-check. We get the same values when we
substitute into πΏ sub two. As expected, we do indeed get the
value of six, seven, zero. And of course, if we had
substituted π‘ sub one equals one and π‘ sub two equal zero into that third equation
that we didnβt use, we would have expected this result. So since the direction vectors of
the two lines are perpendicular and they intersect at a point, we can say that the
lines themselves must also be perpendicular.