Question:
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+PB2= k2, where k is a constant.
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+PB2= k2, where k is a constant.
Updated On: Oct 21, 2023
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Solution and Explanation
The coordinates of points A and B are given as (3, 4, 5) and (-1, 3, -7) respectively.
Let the coordinates of point P be (x, y, z).
On using distance formula, we obtain
PA2 = (x-3)2+(y-4)2+(z−5)2
= x2+9-6x+ y2+16-8y+z2+25-10z
= x2-6x +y2-8y + z2-10z
PB2 = (x+1)2+(y-3)2+(z+7)2
= x2+2x+ y2−6y+z2+14z+59
Now, if PA2+PB2=k2, then
(x2-6x+ y2-8y+z2-10z+50)+(x2+2x+ y2 −6y+z2+14z+59) = k2
⇒2x2+2y2+2z2-4x-14y+4z+109= k2
⇒2(x2+ y2+z2-2x-7y+2z)=k2-109
⇒x2 + y2+ z2 -2x -7y +2z =\(\frac{k^2-109}{2}\)
Thus, the required equation is x2+y2+z2-2x -7y +2z = \(\frac{k^2-109}{2}\)
Let the coordinates of point P be (x, y, z).
On using distance formula, we obtain
PA2 = (x-3)2+(y-4)2+(z−5)2
= x2+9-6x+ y2+16-8y+z2+25-10z
= x2-6x +y2-8y + z2-10z
PB2 = (x+1)2+(y-3)2+(z+7)2
= x2+2x+ y2−6y+z2+14z+59
Now, if PA2+PB2=k2, then
(x2-6x+ y2-8y+z2-10z+50)+(x2+2x+ y2 −6y+z2+14z+59) = k2
⇒2x2+2y2+2z2-4x-14y+4z+109= k2
⇒2(x2+ y2+z2-2x-7y+2z)=k2-109
⇒x2 + y2+ z2 -2x -7y +2z =\(\frac{k^2-109}{2}\)
Thus, the required equation is x2+y2+z2-2x -7y +2z = \(\frac{k^2-109}{2}\)
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