If (2, 3, 5) is one end of a diameter of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of the given sphere is $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2

Vedclass · Accepted Answer

$(4, 9, -3)$

Vedclass · Answer

(D) The equation of the given sphere is $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get $2u = -6, 2v = -12, 2w = -2$.Thus,$u = -3, v = -6, w = -1$.The center of the sphere is $(-u, -v, -w) = (3, 6, 1)$.Let the given end of the diameter be $A = (2, 3, 5)$ and the other end be $B = (\alpha, \beta, \gamma)$.Since the center of the sphere is the midpoint of the diameter,we have:$\frac{\alpha + 2}{2} = 3 \Rightarrow \alpha + 2 = 6 \Rightarrow \alpha = 4$$\frac{\beta + 3}{2} = 6 \Rightarrow \beta + 3 = 12 \Rightarrow \beta = 9$$\frac{\gamma + 5}{2} = 1 \Rightarrow \gamma + 5 = 2 \Rightarrow \gamma = -3$Therefore,the coordinates of the other end are $(4, 9, -3)$.

If $(2, 3, 5)$ is one end of a diameter of the sphere $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$,then the coordinates of the other end of the diameter are:

Similar Questions

The centre of the sphere $\alpha \,r^2 - 2u \cdot r = \beta ,(\alpha \ne 0)$ is

How many different spheres of radius $r$ can be drawn which touch all the three coordinate axes?

If two spheres of radii $r_1$ and $r_2$ cut orthogonally,then the radius of the common circle is

$A$ point moves such that the sum of its distances from the points $(4, 0, 0)$ and $(-4, 0, 0)$ remains $10$. The locus of the point is

The shortest distance from the plane $12x + 4y + 3z = 327$ to the sphere $x^2 + y^2 + z^2 + 4x - 2y - 6z = 155$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module