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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΅?
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In this question, weβre given some information about a sphere.
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First, weβre told the standard equation of the sphere.
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Weβre also told that the line segment π΄π΅ is a diameter of our sphere and weβre told the coordinates at the point π΄.
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We need to use all of this information to determine the coordinates of point π΅.
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Thereβs several different methods we could do this.
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However, usually in problems like this, the easiest method involves starting by writing down all of the information weβre given.
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To do this, letβs start by recalling the standard form for the equation of a sphere.
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We recall a sphere of radius π centered at the point π, π, π will have the following equation in standard form.
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π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared.
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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.
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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.
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Thereβs two different methods of finding the center point.
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We could rewrite the expressions inside of our parentheses as the variable minus some constant.
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However, we could also just find the value of the variable which makes this term equal to zero.
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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.
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Either method would work; itβs personal preference which one you want to use.
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Either way, weβve shown the center of the sphere given to us in the question is the point negative two, negative one, one.
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Similarly, we can find the radius of this sphere by taking the square root of 38.
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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.
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And we might want to sketch this information on a sphere.
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However, itβs not necessary.
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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.
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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.
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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.
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And we can combine all of this information to find the coordinates of point π΅.
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First, line segment π΄πΆ and line segment πΆπ΅ are both radii of our sphere.
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Theyβre both going to have length π.
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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.
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And we can also see something interesting.
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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.
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So letβs use this to find the coordinates of π΅.
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First, weβre going to need to find the vector ππ.
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And to do this, we need to take the vector ππ and subtract the vector ππ.
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This gives us the following expression, and we can subtract this component-wise.
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This gives us the vector ππ is the vector negative two, negative five, negative three.
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We can then add this onto our sketch.
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And remember, the vector from πΆ to π΅ is also equal to the vector from π΄ to πΆ.
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Now we can find the coordinates of π΅ by adding our vector ππ to the vector ππ.
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And all weβre saying here is we can get to the point π΅ from the center by moving along the vector ππ.
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This gives us the vector ππ will be the vector negative two, negative one, one plus the vector negative two, negative five, negative three.
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And we add these together component-wise to get the vector ππ is the vector negative four, negative six, negative two.
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But remember, the question is not asking us for the vector ππ.
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Itβs asking us for the coordinates of the point π΅.
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And of course, the coordinates of the point π΅ will just be the components of our vector π.
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And this gives us that π΅ is the point negative four, negative six, negative two.
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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.