Question Video: Finding the Coordinates of the Other Point on the Diameter of a Sphere Mathematics

Given that 𝐴 (0, 4, 4), and that line segment 𝐴𝐡 is a diameter of the sphere (π‘₯ + 2)Β² + (𝑦 + 1)Β² + (𝑧 βˆ’ 1)Β² = 38, what is the point 𝐡?

03:50

Video Transcript

Given that 𝐴 is the point zero, four, four and that the line segment 𝐴𝐡 is a diameter of the sphere π‘₯ plus two all squared plus 𝑦 plus one all squared plus 𝑧 minus one all squared is equal to 38, what is the point 𝐡?

In this question, we’re given some information about a sphere. First, we’re told the standard equation of the sphere. We’re also told that the line segment 𝐴𝐡 is a diameter of our sphere and we’re told the coordinates at the point 𝐴. We need to use all of this information to determine the coordinates of point 𝐡. There’s 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 we’re given. To do this, let’s start by recalling the standard form for the equation of a sphere. We recall a sphere of radius π‘Ÿ centered at the point π‘Ž, 𝑏, 𝑐 will have the following equation in standard form. π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to π‘Ÿ squared.

This means if we’re given the equation of a sphere in standard form, we can find its center point π‘Ž, 𝑏, 𝑐 and we can also find its radius π‘Ÿ. 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. There’s 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‘Ž would be negative two, our value of 𝑏 would be negative one, and our value of 𝑐 would be one. Either method would work; it’s personal preference which one you want to use. Either way, we’ve 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 we’re going to want to do is use the fact that the line segment 𝐴𝐡 is a diameter of our sphere and that the coordinates of the point 𝐴 are zero, four, four. And we might want to sketch this information on a sphere. However, it’s not necessary. We can actually just do this on a circle because if the line segment 𝐴𝐡 is a diameter of the sphere, then it’s also a diameter of the circle of the same radius. In either case, the only information we need is the line segment 𝐴𝐡 is a diameter of our sphere, so it’s a straight line passing through the center of our sphere. And we know the coordinates of the point 𝐴 is zero, four, four and the coordinates of our center 𝑐 is negative two, negative one, one.

And we can combine all of this information to find the coordinates of point 𝐡. First, line segment 𝐴𝐢 and line segment 𝐢𝐡 are both radii of our sphere. They’re both going to have length π‘Ÿ. Next, because we know the coordinates of the point 𝐴 and the coordinates of the point 𝐢, we’re able to find the vector from 𝐴 to 𝐢. And we can also see something interesting. This is going to be exactly the same as the vector from 𝐢 to 𝐡 because they have the same magnitude of π‘Ÿ and they point in the same direction. So let’s use this to find the coordinates of 𝐡. First, we’re going to need to find the vector 𝐀𝐂. And to do this, we need to take the vector πŽπ‚ and subtract the vector πŽπ€. This gives us the following expression, and we can subtract this component-wise. This gives us the vector 𝐀𝐂 is the vector negative two, negative five, negative three. We can then add this onto our sketch. And remember, the vector from 𝐢 to 𝐡 is also equal to the vector from 𝐴 to 𝐢.

Now we can find the coordinates of 𝐡 by adding our vector πŽπ‚ to the vector 𝐀𝐂. And all we’re saying here is we can get to the point 𝐡 from the center by moving along the vector 𝐀𝐂. This gives us the vector 𝐎𝐁 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 𝐎𝐁 is the vector negative four, negative six, negative two. But remember, the question is not asking us for the vector 𝐎𝐁. It’s asking us for the coordinates of the point 𝐡. And of course, the coordinates of the point 𝐡 will just be the components of our vector 𝐁. And this gives us that 𝐡 is the point negative four, negative six, negative two.

Therefore, we were able to show if 𝐴 is the point zero, four, four and the line segment 𝐴𝐡 is a diameter of the sphere π‘₯ plus two all squared plus 𝑦 plus one all squared plus 𝑧 minus one all squared is equal to 38, then the point 𝐡 must have coordinates negative four, negative six, negative two.

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