Find the point on the sphere x2 + y2 * z2 :4 which is farthest from the point (1, -1, 1),?

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1 Answer
Dec 20, 2017

#(-2sqrt3/3, 2sqrt3/3, -2sqrt3/3)#

Explanation:

Microsoft Paint

Above, I have drawn a picture of a sphere with center C, and a random point P. I have also drawn a line segment connecting P to a point Q on the circle. From P and Q, we can construct two more segments meeting at C forming a triangle. The length of QC and PC will be constant no matter what point Q we choose.

Therefore, to maximize d, we must choose the largest possible angle between the two other legs of the triangle, which is #180^@#, in which case, d will pass through the center point, C.

The sphere given in the question is centered on the origin.

#x^2 + y^2 + z^2 = 4#

The point given has the same magnitude in each dimension.

|1| = |-1| = |1|

Since they must be co-linear, the desired farthest point must also have the same magnitude in each dimension. In other words;

#|x|=|y|=|z|#

We can therefore rewrite the function for our sphere as;

#x^2 + x^2 + x^2 = 4#

And solve for x.

#3x^2 = 4#

#x^2 = 4/3#

#|x|=2sqrt3/3#

Because the sphere is centered on the origin, we can assume that the sign for the x, y, and z values of this point will be opposite of those of point P. Our point is therefore,

#(-2sqrt3/3, 2sqrt3/3, -2sqrt3/3)#