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6. If A and B be the points (3, 4, 5) and (-1, 3, -7), respectively, find the equation of the set of points P such that PA ^2 + PB ^ 2 = k^ 2 , where k is a constant.

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Given points,

A (3, 4, 5) and B  (-1, 3, -7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

PB^2=(x-(-1))^2+(y-(3))^2+(z-(-7))^2

PB^2=x^2+2x+1+y^2-6y+9+z^2+14z+49

PB^2=x^2+y^2+z^2+2x-6y+14z+59

And

PA^2=(x-3)^2+(y-4)^2+(z-5)^2

PA^2=x^2+y^2+z^2-6x-8y-10z+50

Now, given Condition

PA ^2 + PB ^ 2 = k^ 2

x^2+y^2+z^2+2x-6y+14z+59+x^2+y^2+z^2-6x-8y-10z+50=k^2

2x^2+2y^2+2z^2-4x-14y+4z+109=k^2

Hence, equation of the set of points P is 

2x^2+2y^2+2z^2-4x-14y+4z+109=k^2.

Posted by

Pankaj Sanodiya

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