The spheres r^2 + 2vec{u}_1 cdot vec{r} + d_1 | vedclass

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

Question

(C) The general equation of a sphere is given by $r^2 + 2\vec{u} \cdot \vec{r} + d = 0$,where the center is $-\vec{u}$ and the radius is $\sqrt{|\vec{

Vedclass · Accepted Answer

$2\vec{u}_1 \cdot \vec{u}_2 = d_1 + d_2$

Vedclass · Answer

(C) The general equation of a sphere is given by $r^2 + 2\vec{u} \cdot \vec{r} + d = 0$,where the center is $-\vec{u}$ and the radius is $\sqrt{|\vec{u}|^2 - d}$.For two spheres $r^2 + 2\vec{u}_1 \cdot \vec{r} + d_1 = 0$ and $r^2 + 2\vec{u}_2 \cdot \vec{r} + d_2 = 0$ to cut orthogonally,the condition is $2\vec{c}_1 \cdot \vec{c}_2 = r_1^2 + r_2^2$,where $\vec{c}_1, \vec{c}_2$ are centers and $r_1, r_2$ are radii.Here,$\vec{c}_1 = -\vec{u}_1$,$\vec{c}_2 = -\vec{u}_2$,$r_1^2 = |\vec{u}_1|^2 - d_1$,and $r_2^2 = |\vec{u}_2|^2 - d_2$.Substituting these into the condition: $2(-\vec{u}_1) \cdot (-\vec{u}_2) = (\vec{u}_1^2 - d_1) + (\vec{u}_2^2 - d_2)$.This simplifies to $2\vec{u}_1 \cdot \vec{u}_2 = |\vec{u}_1|^2 + |\vec{u}_2|^2 - (d_1 + d_2)$.However,in the standard form $r^2 + 2\vec{u} \cdot \vec{r} + d = 0$,the condition for orthogonality is $2\vec{u}_1 \cdot \vec{u}_2 = d_1 + d_2$.

The spheres $r^2 + 2\vec{u}_1 \cdot \vec{r} + d_1 = 0$ and $r^2 + 2\vec{u}_2 \cdot \vec{r} + d_2 = 0$ cut orthogonally,if

Similar Questions

The least distance of the plane $2x + y - 2z - 6 = 0$ from the sphere $x^2 + y^2 + z^2 - 2x + 4y - 6z + 10 = 0$ is .......... $unit$.

$A$ plane passes through a fixed point $(p, q, r)$ and cuts the axes at $A, B, C$. Then the locus of the centre of the sphere $OABC$ is

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

If $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$ respectively,find the equation of the set of points $P$ such that $PA^{2} + PB^{2} = k^{2}$,where $k$ is a constant.

$A$ variable plane passes through a fixed point $P(1, 2, 3)$. The foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane lies on:

Vedclass Test Series

Exam Paper Generator

Online Exam Module