### Video Transcript

Find, to the nearest hundredth, the
distance between the parallel lines πΏ one, π₯ plus seven over nine equals π¦ plus
one over five equals π§ minus seven over negative six, and πΏ two, π₯ plus three
over nine equals π¦ plus 10 over five equals π§ plus 10 over negative six.

Okay, so here we have these two
lines that are parallel, πΏ one and πΏ two. Our question asks us to solve for
the distance between them, meaning the minimum or perpendicular distance between
these lines. Letβs call that distance π, and we
can recall the mathematical relationship for this distance. What weβll need to know are the
components of a vector, weβve called it π¬, that runs parallel to our two lines. And weβll also need to know the
components of a vector ππ that runs from a point on our second line to a point on
our first line. We can say then that, to calculate
π, the coordinates of points π and πΏ and the components of vector π¬ are what we
need to know.

To solve for this information,
letβs look at the given equations of lines one and two. Starting with line one, this is
given in whatβs called symmetric form. We can write these three fractions
as being equal to one another because we say that theyβre also equal to a scale
factor. We can call it π‘ one. The fact that these three fractions
are all equal to π‘ one means that we can write separate equations for the π₯-, π¦-,
and π§-coordinates of line πΏ one. That is, since π₯ plus seven
divided by nine equals π‘ one, we can write that π₯ equals nine times π‘ one minus
seven. In a similar way, since π¦ plus one
divided by five equals π‘ one, π¦ is equal to five times π‘ one minus one. And then, π§ minus seven over
negative six equaling π‘ one implies that π§ equals negative six π‘ one plus
seven.

Line πΏ one is now written in
whatβs called parametric form. We can combine these three separate
equations into one vector equation. The vector with components π₯, π¦,
π§ is equal to the vector to negative seven, negative one, seven plus our scale
factor π‘ one times the vector nine, five, negative six.

Now, if we take our π₯-, π¦-, and
π§-components and write them as the vector π«, then this equals the vector negative
seven, negative one, seven plus π‘ one times the vector nine, five, negative
six. This first vector is a vector from
the origin of a coordinate system to a point on line πΏ one. That is, the point with coordinates
negative seven, negative one, seven lies on πΏ one. And then, regarding the second
vector, this is a vector that lies along the axis of this line. That means itβs parallel to the
line. And therefore, we can say that this
is our vector π¬.

So far then, we have the components
of a vector parallel to both of our lines. And we also have the coordinates of
a point on our first line. To find the coordinates of a point
on our second line, letβs look at that lineβs equation. Once again, this equation is given
to us in symmetric form. That means we can say that each one
of these fractions is equal to another scale factor weβll call π‘ two. If we once again write out separate
equations for π₯, π¦, and π§, we find that π₯ equals nine π‘ two minus three, π¦
equals five π‘ two minus 10, and π§ equals negative six π‘ two minus 10.

Once again, we can write these
parametric equations as a vector equation. A vector with components π₯, π¦,
and π§ equals the vector negative three, negative 10, negative 10 plus π‘ two times
the vector nine, five, negative six. Written this way, we know that this
vector travels from the origin of a coordinate frame to a point along our line πΏ
two. This tells us that the coordinates
of that point are simply the components of the vector. We can write then that the
coordinates of the point weβve called π are negative three, negative 10, negative
10. Our next step is to use the
coordinates of points πΏ and π to solve for the vector ππ. This is the vector with components
equal to the difference between the coordinates of point πΏ and those of point
π. ππ, therefore, has components
negative four, nine, 17.

We can now move ahead by
calculating the cross product of vectors ππ and π¬. That equals the determinant of this
matrix. And notice that, by column, we have
the π₯-, π¦-, and π§-components of our two vectors. Calculating this cross product, we
get π’ hat times negative 139 minus π£ hat times negative 129 plus π€ hat times
negative 101. We can write this result then as a
vector with components negative 139, positive 129, and negative 101. Now that we know ππ cross π¬,
weβre ready to calculate the magnitude of this cross product and divide it by the
magnitude of π¬.

The magnitude of ππ cross π¬
equals the square root of negative 139 squared plus 129 squared plus negative 101
squared. We divide this by the magnitude of
π¬, which itself is the square root of nine squared plus five squared plus negative
six squared. Calculating this entire fraction
altogether, rounding to the nearest hundredth, we get an answer of 18.03. We say then that the distance
between these two parallel lines is 18.03 length units.