If the plane 2ax - 3ay + 4az + 6 = 0 passes t | vedclass

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

Question

(A) The equation of the first sphere is ${S_1} \equiv {x^2} + {y^2} + {z^2} + 6x - 8y - 2z - 13 = 0$. Its center ${C_1}$ is $(-3, 4, 1)$.The equation

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(A) The equation of the first sphere is ${S_1} \equiv {x^2} + {y^2} + {z^2} + 6x - 8y - 2z - 13 = 0$. Its center ${C_1}$ is $(-3, 4, 1)$.The equation of the second sphere is ${S_2} \equiv {x^2} + {y^2} + {z^2} - 10x + 4y - 2z - 8 = 0$. Its center ${C_2}$ is $(5, -2, 1)$.The midpoint $P$ of the line segment joining ${C_1}$ and ${C_2}$ is given by:$P = \left( \frac{-3 + 5}{2}, \frac{4 - 2}{2}, \frac{1 + 1}{2} \right) = (1, 1, 1)$.Since the plane $2ax - 3ay + 4az + 6 = 0$ passes through the point $P(1, 1, 1)$,we substitute the coordinates of $P$ into the plane equation:$2a(1) - 3a(1) + 4a(1) + 6 = 0$$2a - 3a + 4a + 6 = 0$$3a + 6 = 0$$3a = -6$$a = -2$.

If the plane $2ax - 3ay + 4az + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres ${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$ and ${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$,then $a$ equals

Similar Questions

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:

$A$ 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

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:

The radius of the circular section of the sphere $|r| = 5$ by the plane $r \cdot (i + j + k) = 3\sqrt{3}$ is

The radius of the circle formed by the intersection of the sphere $x^2+y^2+z^2+2x-2y-4z-19=0$ and the plane $x+2y+2z+7=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module