P is a variable point and the co-ordinates of two points A and B are -2-2-3 and 13-3-13 respectively- If 3 P A-2 P B- then locus of P is -A- x 2- y 2- z 2-28 x -12 y -10 z -247-0B- x2-y2-z2-28 x-12 y-10 z-147-0- x2-y2-z2-28 x-12 y-10 z-147-0D- x 2- y 2- z 2-28 x -12 y -10 z -247-0

The correct option is A x^2 + y^2 + z^2 + 28x 12y + 10z 247 = 0 3PA = 2PB ⇒ 9PA^2 = 4PB^2 ⇒ 9{ (x+2)^2+(y 2)^2+(z 3)^2 }=4{ (x 13)2+(y 3)^2+(z 13)^2 } ⇒ 5x ...

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Standard XII

Mathematics

Distance Formula

P is a variab...

Question

P is a variable point and the co-ordinates of two points A and B are (–2, 2, 3) and (13, –3, 13) respectively. If 3PA = 2PB, then locus of P is :

x^{2} + y^{2} + z^{2} - 28 x + 12 y + 10 z - 247 = 0

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x^{2} + y^{2} + z^{2} + 28 x - 12 y + 10 z - 247 = 0

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x^{2} + y^{2} + z^{2} + 28 x - 12 y + 10 z - 147 = 0

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x^{2} + y^{2} + z^{2} - 28 x - 12 y + 10 z + 147 = 0

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Solution

The correct option is B $x^{2} + y^{2} + z^{2} + 28 x - 12 y + 10 z - 247 = 0$
$3 P A = 2 P B \Rightarrow 9 P A^{2} = 4 P B^{2} \Rightarrow 9 {(x + 2)^{2} + (y - 2)^{2} + (z - 3)^{2}} = 4 {(x - 13) 2 + (y - 3)^{2} + (z - 13)^{2}} \Rightarrow 5 x^{2} + 5 y^{2} + 5 z^{2} + 140 x - 60 y + 50 z - 1235 = 0 \Rightarrow x^{2} + y^{2} + z^{2} + 28 x - 12 y + 10 z - 247 = 0$

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P is a variable point and the co-ordinates of two points A and B are (–2, 2, 3) and (13, –3, 13) respectively. If 3PA = 2PB, then locus of P is :

The correct option is B x2+y2+z2+28x−12y+10z−247=0 3PA=2PB⇒9PA2=4PB2⇒9{(x+2)2+(y−2)2+(z−3)2}=4{(x−13)2+(y−3)2+(z−13)2}⇒5x2+5y2+5z2+140x−60y+50z−1235=0⇒x2+y2+z2+28x−12y+10z−247=0