Question

Answer this Question of iit jee . Chapter → Three dimensional Geometry​

Answer 1

Answer:

let the other end of the diameter be (p,q,r).

The general form of sphere is x²+y²+z²+2ax+2by+2cz+d=0

where (-a,-b,-c) is the centre of the sphere.

So, the centre of the sphere x²+y²+z²-6x-12y-2z+20 = 0 is (3,6,1).

we know , centre is the midpoint of the diameter.

$→3 = \frac{p + 2}{2}$

$→6 = p + 2$

$∴p = 4$

and

$6 = \frac{q + 3}{2}$

$→12 = q + 3$

$∴q = 9$

and

$1 = \frac{r + 5}{2}$

$→2 = r + 5$

$∴r = - 3$

Hence, the coordinates of the other end of Diameter is (4,9,-3).

Answer 2

$\large\underline{\sf{Solution-}}$

Given the equation of sphere as

$\rm :\longmapsto\: {x}^{2}+{y}^{2} + {z}^{2} - 6x - 12y - 2z + 20 = 0$

and

$\rm :\longmapsto\:one \: end \: of \: diameter \: as \: (2,3,5)$

Let assume that AB be the diameter of given sphere.

And

Let coordinates of A be (2, 3, 5).

So, we have to find the coordinates of B.

Let O be the center of sphere.

We know, Center is the midpoint of diameter.

So, O is the midpoint of AB.

So, Let first find coordinates of O.

We know,

The center of sphere x²+y²+z²+2ux+2vy+2wz+d=0 is given by

$\rm \: = \: \: \bigg(\dfrac{ - 1}{2}coeff. \: of \: x,\dfrac{ - 1}{2}coeff. \: of \: y, \dfrac{ - 1}{2}coeff. \: of \: z \bigg)$

$\rm \: = \: \: ( - u, - v, - w)$

So,

Given equation of sphere is

$\rm :\longmapsto\: {x}^{2}+{y}^{2} + {z}^{2} - 6x - 12y - 2z + 20 = 0$

So,

⟼ Coordinates of center, O = ( 3, 6, 1 ).

Now, we have

⟼ Coordinates of A = ( 2, 3, 5 )

⟼ Coordinates of O = ( 3, 6, 1 )

Now, let assume that

⟼ Coordinates of B be ( a, b, c ).

So, by using Midpoint Formula, we get

$\rm :\longmapsto\:(3,6,1) = \bigg(\dfrac{a + 2}{2},\dfrac{b + 3}{2}, \dfrac{c + 5}{2} \bigg)$

$\rm :\longmapsto\:\dfrac{a + 2}{2} = 3, \: \dfrac{b + 3}{2} =6, \: \dfrac{c + 5}{2} = 1$

$\rm :\longmapsto\:a + 2 = 6, \: b + 3 = 12, \: c + 5 = 2$

$\rm :\longmapsto\:a = 4, \: b = 9, \: c= - 3$

Hence,

Coordinates of B = ( 4, 9, - 3 )

So, Option (a) is correct