Question

# Find the locus of the point P if AP2−BP2=18, where A ≡ (1, 2, –3) and B ≡ (3, –2, 1)

A

2x + 3y + 4 = 9

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B

2x + 4y + 4z = 9

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C

2x – 3y + 4z = 9

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D

2x – 4y + 4z = 9

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Solution

## The correct option is B 2x – 4y + 4z = 9 We want to find the locus of P. That is, set of Ps satisfying the given condition. For that, let the point P be (x, y, z). Using distance formula, we get AP2=(x−1)2+(y−2)2+(z+3)2 and BP2=(x−3)2+(y+2)2+(z−1)2 Given that AP2−BP2=18 ⇒(x−1)2+(y−2)2+(z+3)2−{(x−3)2+(y+2)2+(z−−1)2}=18 ⇒x2+1−2x+y2+4−4y+z2+9+6z−{x2+9−6x+y2+4+4y+z2+1−2z} ⇒x2+y2+z2−2x−4y+6z+14−x2−y2−z2+6x−4y+2z−14=28 ⇒ 4x–8y+8z=18 ⇒ 2x–4y+4z=9 Locus of the point P is 2x – 4y + 4z = 9

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