Determine the points in i. xy-plan e ii. yz-plane and iii zx-plane which re equidistant from the points `A(1,-1,0),\ B(2,1,2),\ a n d\ C(3,2,-1` )

Explanation

Given that the purpose A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)
(i) XY plane :
We know z = zero in xy-plane.
So let P(x, y, 0) be any purpose in xy-plane consistent with the question:
`PA = PB = PC`
`PA^2 = PB^2 = PC^2 `
By using the formula, The distance between any two points (a, b, c) and (k, l, m) is given by,
`sqrt((a-k)^2+(b-l)^2+(c-m)^2)`
so,the distance between P and A is PA
`=sqrt((x-1)^2+(y+1)^2+(0-0)^2)`
`=sqrt((x-1)^2+(y+1)^2)`
the distance between P and B is PB
`=sqrt((x-2)^2+(y-1)^2+(0-2)^2)`
`=sqrt((x-2)^2+(y-1)^2+4)`
the distance between P and C is PC
`=sqrt((x-3)^2+(y-2)^2+(0+1)^2)`
`=sqrt((x-3)^2+(y-2)^2+1)`
We know that `PA^2 = PB^2`
so,`(x-1)^2+(y+1)^2=(x-2)^2+(y-1)^2+4`
By solving this we get,
`2x=-4y+7`......(1)
Again `PA^2 = PC^2`
so,`(x-1)^2+(y+1)^2=(x-3)^2+(y-2)^2+1`
by solving this we get, `2x+3y-6=0`
Now put the value of 2x from (i) we get,
`7-4y+3y-6=0`
`y=1`
now substituting the value of y in (i) we get,
`2x=7-4`
`x=frac{3}{2}`
thereforeThe purpose `P (3/2, 1, 0)` in xy-plane is equidistant from A, B and C.
Similarly The purpose `Q (0, 31/16, -3/16)` in yz-plane is equidistant from A, B and C.
and The purpose `R (31/10, 0, 1/5)` in xz-plane is equidistant from A, B and C.