The intersection of the spheres {x^2} + {y^2} | vedclass

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

Question

(A) The equation of the plane passing through the intersection of two spheres ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.Given the sp

Vedclass · Accepted Answer

$2x - y - z = 1$

Vedclass · Answer

(A) The equation of the plane passing through the intersection of two spheres ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.Given the spheres:${S_1}: {x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$${S_2}: {x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$Subtracting the two equations:$({x^2} + {y^2} + {z^2} + 7x - 2y - z - 13) - ({x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8) = 0$$(7x - (-3x)) + (-2y - 3y) + (-z - 4z) + (-13 - (-8)) = 0$$10x - 5y - 5z - 5 = 0$Dividing the entire equation by $5$:$2x - y - z = 1$Thus,the intersection of the two spheres lies on the plane $2x - y - z = 1$.

The intersection of the spheres ${x^2} + {y^2} + {z^2} + 7x - 2y - z = 13$ and ${x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8$ is the same as the intersection of one of the spheres and the plane:

Similar Questions

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.

The radius of the circle formed by the intersection of the plane $x + 2y + 2z = 15$ and the sphere ${x^2} + {y^2} + {z^2} - 2y - 4z = 11$ is

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

The equation of the sphere passing through the points $(1,0,0), (0,1,0)$ and $(1,1,1)$ and having the smallest radius is:

If $r$ is the position vector of any point on a sphere and $a$ and $b$ are the position vectors of the extremities of a diameter,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module