### Video Transcript

Given that π΄ is the point zero,
four, four and that the line segment π΄π΅ is a diameter of the sphere π₯ plus two
all squared plus π¦ plus one all squared plus π§ minus one all squared is equal to
38, what is the point π΅?

In this question, weβre given some
information about a sphere. First, weβre told the standard
equation of the sphere. Weβre also told that the line
segment π΄π΅ is a diameter of our sphere and weβre told the coordinates at the point
π΄. We need to use all of this
information to determine the coordinates of point π΅. Thereβs several different methods
we could do this. However, usually in problems like
this, the easiest method involves starting by writing down all of the information
weβre given. To do this, letβs start by
recalling the standard form for the equation of a sphere. We recall a sphere of radius π
centered at the point π, π, π will have the following equation in standard
form. π₯ minus π all squared plus π¦
minus π all squared plus π§ minus π all squared is equal to π squared.

This means if weβre given the
equation of a sphere in standard form, we can find its center point π, π, π and
we can also find its radius π. And we could see the equation given
to us in the question is in standard form, so we can use this to find the center and
radius of our sphere. Thereβs two different methods of
finding the center point. We could rewrite the expressions
inside of our parentheses as the variable minus some constant. However, we could also just find
the value of the variable which makes this term equal to zero. So, for example, our value of π
would be negative two, our value of π would be negative one, and our value of π
would be one. Either method would work; itβs
personal preference which one you want to use. Either way, weβve shown the center
of the sphere given to us in the question is the point negative two, negative one,
one.

Similarly, we can find the radius
of this sphere by taking the square root of 38. The last thing weβre going to want
to do is use the fact that the line segment π΄π΅ is a diameter of our sphere and
that the coordinates of the point π΄ are zero, four, four. And we might want to sketch this
information on a sphere. However, itβs not necessary. We can actually just do this on a
circle because if the line segment π΄π΅ is a diameter of the sphere, then itβs also
a diameter of the circle of the same radius. In either case, the only
information we need is the line segment π΄π΅ is a diameter of our sphere, so itβs a
straight line passing through the center of our sphere. And we know the coordinates of the
point π΄ is zero, four, four and the coordinates of our center π is negative two,
negative one, one.

And we can combine all of this
information to find the coordinates of point π΅. First, line segment π΄πΆ and line
segment πΆπ΅ are both radii of our sphere. Theyβre both going to have length
π. Next, because we know the
coordinates of the point π΄ and the coordinates of the point πΆ, weβre able to find
the vector from π΄ to πΆ. And we can also see something
interesting. This is going to be exactly the
same as the vector from πΆ to π΅ because they have the same magnitude of π and they
point in the same direction. So letβs use this to find the
coordinates of π΅. First, weβre going to need to find
the vector ππ. And to do this, we need to take the
vector ππ and subtract the vector ππ. This gives us the following
expression, and we can subtract this component-wise. This gives us the vector ππ is
the vector negative two, negative five, negative three. We can then add this onto our
sketch. And remember, the vector from πΆ to
π΅ is also equal to the vector from π΄ to πΆ.

Now we can find the coordinates of
π΅ by adding our vector ππ to the vector ππ. And all weβre saying here is we can
get to the point π΅ from the center by moving along the vector ππ. This gives us the vector ππ will
be the vector negative two, negative one, one plus the vector negative two, negative
five, negative three. And we add these together
component-wise to get the vector ππ is the vector negative four, negative six,
negative two. But remember, the question is not
asking us for the vector ππ. Itβs asking us for the coordinates
of the point π΅. And of course, the coordinates of
the point π΅ will just be the components of our vector π. And this gives us that π΅ is the
point negative four, negative six, negative two.

Therefore, we were able to show if
π΄ is the point zero, four, four and the line segment π΄π΅ is a diameter of the
sphere π₯ plus two all squared plus π¦ plus one all squared plus π§ minus one all
squared is equal to 38, then the point π΅ must have coordinates negative four,
negative six, negative two.