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NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions

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Question : 20 of 20
Marks: +1, -0
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 PA2+PB2PA^2 + PB^2 = k2k^2, where k is a constant.
Solution:  
Let P(x, y, z) be any point
Then PA = (x−3)2+(y−4)2+(z−5)2\sqrt{(x-3)^2+(y-4)^2+(z-5)^2}
=
x2+9−6x+y2+16−8y+z2+25−10z\sqrt{x^2+9-6x+y^2+16-8y+z^2+25-10z}
PB = (x+1)2+(y−3)2+(z+7)2\sqrt{(x+1)^2+(y-3)^2+(z+7)^2}
=
x2+1+2x+y2+9−6y+z2+49+14z\sqrt{x^2+1+2x+y^2+9-6y+z^2+49+14z}
Now, PA2+PB2PA^2 + PB^2 = k2k^2
∴
[x2+9−6x+y2+16−8y+z2+25−10z]2[\sqrt{x^2+9-6x+y^2+16-8y+z^2+25-10z}]^2
+
[x2+1+2x+y2+9−6y+z2+49+14z]2[\sqrt{x^2+1+2x+y^2+9-6y+z^2+49+14z}]^2
= k2k^2
∴ x2x^2 + 9 - 6x + y2y^2 + 16 - 8y + z2z^2 + 25 - 10z + x2x^2 + 1 + 2x + y2y^2 - 9 - 6y + z2z^2 + 49 + 14z = k2k^2
⇒ 2x2+2y2+2z22x^2+2y^2+2z^2 - 4x - 14y + 4z + 109 = k2k^2
⇒ 2 (x2+y2+z2−2z−7y+2z)(x^2+y^2+z^2 -2z-7y+2z) = k2k^2 - 109
⇒ x2+y2+z2x^2+y^2+z^2 - 2x - 7y + 2z = k2−1092\frac{k^2-109}{2}
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