### Video Transcript

Find to the nearest hundredth the
distance between the two planes vector six, three, six dot vector π« equals 11 and
π₯ over three plus π¦ over six plus π§ over three equals one.

In this question, we want to find
the distance between two planes. And to do this, we start by
checking whether the two planes are parallel. We can recall that two planes are
parallel if the normal vectors to each plane are parallel. The normal vector to the plane
vector six, three, six dot vector π« equals 11 is given by the vector six, three,
six. The normal vector to the other
plane can be found by the coefficients of π₯, π¦, and π§. It can be helpful to rearrange this
equation by multiplying through by six. The normal vector can therefore be
given as the vector two, one, two.

Two vectors are parallel if they
are scalar multiples of each other, and this is what we have here. So that means that the normal
vectors are parallel and indeed then the planes themselves are parallel. So when it comes to finding the
distance between two parallel planes, we can take a point on one of the planes and
work out the perpendicular distance from that point to the other plane. Letβs take the equation of the
second plane and find a point which lies on this plane.

One way that we can do this is by
substituting in π₯ is equal to zero and π¦ is equal to zero. Doing this gives us the equation
zero over three plus zero over six plus π§ over three equals one. When we simplify this, we get π§
over three equals one, and so π§ is equal to three. Since we set π₯ and π¦ equal to
zero, then we know that the point zero, zero, three lies on this plane. We can then recall and use the
following formula.

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 of π₯ sub one, π¦ sub
one, and π§ sub one will be zero, zero, and three, respectively. The values of π, π, and π come
from the other plane, and they will be six, three, and six, respectively.

Notice, however, that in this form
of the plane, negative π will be given by 11. And so that means that π must be
equal to negative 11. We are now ready to substitute
these values into the formula. So we have π· is equal to the
magnitude of six times zero plus three times zero plus six times three plus negative
11 over the square root of six squared plus three squared plus six squared. Simplifying, we have the magnitude
of 18 minus 11 over the square root of 36 plus nine plus 36.

On the numerator, we have the
magnitude of seven, which is seven. And on the denominator, we have the
square root of 81. We can then write that this
distance must be seven over nine length units. This is a perfectly valid answer,
but notice that we were asked for the value to the nearest hundredth. And so we need to write this as a
decimal. We know that seven-ninths as a
decimal is 0.7 recurring. And therefore, rounded to the
nearest hundredth, we can give the answer that the distance between the two planes
must be 0.78 length units.