### Video Transcript

Find, to the nearest hundredth, the
distance between the two planes π₯ plus two π¦ plus four π§ equals four and two π₯
over 13 plus four π¦ over 13 plus eight π§ over 13 is equal to one.

When we are finding the distance
between two planes, we must first check if 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 each plane is
given by the coefficients of π₯, π¦, and π§. And so the normal vector to the
first given plane is the vector one, two, four. The normal vector to the second
plane can be given as two thirteenths, four thirteenths, eight thirteenths.

Since we can multiply each
component of the first plane by two over 13 to give the corresponding component of
the second plane, then that means that the normal vectors are scalar multiples of
each other, and so they are parallel. Hence, the planes themselves are
also parallel. And so, in order to find the
distance between two planes, we can find a point on one of the planes and then work
out the perpendicular distance from that point to the other plane.

It doesnβt matter which plane we
find a point on. So letβs take the first plane of π₯
plus two π¦ plus four π§ equals four and see if we can find a point on this
plane. We can substitute the values of π₯
equals zero and π¦ equals zero into the equation of this plane. This would give zero plus two times
zero plus four π§ is equal to four. Since four π§ is equal to four,
then we know that π§ is equal to one. We can therefore say that the point
zero, zero, one lies on this first plane.

We can now recall the formula that
the perpendicular distance, denoted uppercase π·, between the point π₯ sub one, π¦
sub one, π§ sub one and the plane ππ₯ plus ππ¦ plus ππ§ plus π equals zero 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 coordinates of the point zero,
zero, one gives us the π₯ sub one, π¦ sub one, and π§ sub one values.

For the coefficients π, π, π,
and π from the plane, we may find it useful to rearrange this equation by
multiplying through by 13. So we have the equation two π₯ plus
four π¦ plus eight π§ minus 13 equals zero. Notice how we have also rearranged
the constant term so that π is on the same side as the equation in the form that we
use in the formula. So weβll be using the values π is
equal to two, π is equal to four, π is equal to eight, and π is equal to negative
13. Now we can substitute these into
the formula.

And so we have π· is equal to the
magnitude of two times zero plus four times zero plus eight times one plus negative
13 over the square root of two squared plus four squared plus eight squared. Simplifying, we have the magnitude
of eight minus 13 over the square root of four plus 16 plus 64. On the numerator, the magnitude of
negative five is five. And on the denominator, we have the
square root of 84.

Because we are asked for a value to
the nearest hundredth, then we need to find a decimal equivalent. And this will be 0.5455 and so on
length units. We can therefore give the answer
that the distance between the two given planes is 0.55 length units to the nearest
hundredth.