### Video Transcript

Find, to the nearest two decimal
places, the distance between the two planes two times π₯ minus two plus π¦ minus
three plus three times π§ minus one equals zero and vector π« dot four, two, six is
equal to 12.

In this question, we are asked to
find the distance between two planes. The first thing we need to do is to
check if the planes are parallel. If they are, we can then apply the
formula for perpendicular distance. We can remember that two planes are
parallel if the normal vectors to each plane are parallel. The first plane is given in a
Cartesian form. When we expand the parentheses, we
can find the coefficients of π₯, π¦, and π§. Using these coefficients, we can
give the normal vector to this plane as the vector two, one, three. The normal vector to the second
plane can be given as the vector four, two, six.

We can then observe that these two
normal vectors are scalar multiples of each other. And that means that the two planes
are parallel. Now that we know that these two
planes are parallel, we can find the distance between them by finding a point which
lies on one of the planes and working out the distance between this point and the
other plane. Letβs find a point which lies on
the first plane. One way to do this is by setting π₯
is equal to zero and π¦ is equal to zero. When we do this, we get the
equation two times negative two plus negative three plus three π§ minus three is
equal to zero. Simplifying this, we get three π§
minus 10 is equal to zero, and so π§ is equal to 10 over three. As we set π₯ and π¦ equal to zero,
we can say that the point zero, zero, 10 over three lies on this plane.

We can then apply the formula that
the perpendicular distance, denoted uppercase π·, between the point π₯ sub one, π¦
sub one, π§ sub one and the plane vector π« dot π, π, π equals negative π is
given by π· equals the magnitude of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one
plus π over the square root of π squared plus π squared plus π squared. The values that we substitute into
the formula can be given as π₯ sub one equals zero, π¦ sub one equals zero, and π§
sub one equals 10 over three. Then, from the plane, we have π is
equal to four, π is equal to two, and π is equal to six. Since the value of negative π is
equal to 12, then we know that π must be equal to negative 12.

We are now ready to substitute
these values into the formula. This will give us π· is equal to
the magnitude of four times zero plus two times zero plus six times 10 over three
plus negative 12 over the square root of four squared plus two squared plus six
squared. We can simplify this to the
magnitude of 20 minus 12 over the square root of 16 plus four plus 36. On the numerator, the magnitude of
eight is eight. And on the denominator, we have the
square root of 56. Since weβre asked to find the
answer to two decimal places, letβs use our calculators to work out a decimal
equivalent. And so we have 1.0690 and so
on. Rounded to two decimal places, we
can give the answer that the distance between the two given planes is 1.07 length
units.

Notice that, in this example, we
found a point on the first given plane and worked out the distance between this
point and the second plane. However, we couldβve done it in the
other direction. For example, we couldβve found a
point on the second plane and worked out the distance between this and the first
plane. Either method would produce the
same perpendicular distance of 1.07 length units to two decimal places.