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

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

Question

(B) The equation of the 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 $u=-3,

Vedclass · Accepted Answer

$(4,9,5)$

Vedclass · Answer

(B) The equation of the 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 $u=-3, v=-6, w=-1$. The centre of the sphere is $(-u,-v,-w) = (3,6,1)$. Let the other end of the diameter be $(\alpha, \beta, \gamma)$. Since the centre of the sphere is the midpoint of the diameter, we have: $(3,6,1) = \left(\frac{\alpha+2}{2}, \frac{\beta+3}{2}, \frac{\gamma-3}{2}\right)$. Equating the coordinates: $\frac{\alpha+2}{2} = 3 \Rightarrow \alpha+2 = 6 \Rightarrow \alpha = 4$. $\frac{\beta+3}{2} = 6 \Rightarrow \beta+3 = 12 \Rightarrow \beta = 9$. $\frac{\gamma-3}{2} = 1 \Rightarrow \gamma-3 = 2 \Rightarrow \gamma = 5$. Thus, the other end of the diameter is $(4,9,5)$.

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

Similar Questions

One mole of potassium dichromate completely oxidises the following number of moles of ferrous sulphate in acidic medium.

Which one of the following is correct?

In which of the following compounds does nitrogen exhibit the highest oxidation state?

Which of the following is non-ionizable?

Which of the following is not optically active?

Vedclass Test Series

Exam Paper Generator

Online Exam Module