$\begingroup$Give a geometric description of the following set of points:
$x^2 +y^2 + z^2-8x+14y-18z>/= 65 $
So I completed the square and got the set to read: $(x-4)^2+(y+7)^2+(z-9)^2>/= 211 $
However the answer in the book says this is the set of all points outside of a ball with center (-4,7,-9) and radius of radical 211. How do you know this is the set of points outside of a ball? What would be different if this was these the set of all points within a ball?
$\endgroup$11 Answer
$\begingroup$$$(x-4)^2+(y+7)^2+(z-9)^2 > 211 $$
Take the square root of both sides.
$$\sqrt{(x-4)^2+(y+7)^2+(z-9)^2} > \sqrt{211} $$
You will recognize the left side as a Euclidian distance. Thus, this expression describes a set of points $(x,y,z)$ that are more than $\sqrt{211}$ in Euclidian distance away from the point $(4,-7,9)$.
A ball is a set of points which are all the same distance away from some point. For instance,
$$\sqrt{(x-4)^2+(y+7)^2+(z-9)^2} = \sqrt{211} $$
is set of points all $\sqrt{211}$ distance away from $(4,-7,9)$. If you wanted all the points inside of this ball, you'd just reverse the inequality:
$$\sqrt{(x-4)^2+(y+7)^2+(z-9)^2} < \sqrt{211} $$
$\endgroup$8
