JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

A

and

B

are the points

(3,4,5)

and

(-1,3,-7)

respectively, then the equation of set of point

P

such that

P A^2+P B^2=k^2

, where

k

is constant, is

View answer and explanation

Correct option: A

2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+109-k^2=0

Option A (

2(x^2+y^2+z^2)-4x-14y+4z+109-k^2=0

) is correct because it represents the equation derived from the given condition

PA^2 + PB^2 = k^2

More Questions from 3D GEOMETRY SYSTEM

Practice All Chapter MCQs →

\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3