The equation |r|^2 - r cdot (2i + 4j - 2k) - | vedclass

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

Question

(C) The general equation of a sphere in vector form is given by $|r|^2 - 2(r \cdot a) + c = 0$,where the radius is $\sqrt{|a|^2 - c}$.Comparing the gi

Vedclass · Accepted Answer

Sphere of radius $4$

Vedclass · Answer

(C) The general equation of a sphere in vector form is given by $|r|^2 - 2(r \cdot a) + c = 0$,where the radius is $\sqrt{|a|^2 - c}$.Comparing the given equation $|r|^2 - r \cdot (2i + 4j - 2k) - 10 = 0$ with the standard form:Here,$2a = (2i + 4j - 2k)$,which implies $a = (i + 2j - k)$.Also,$c = -10$.The radius of the sphere is $\sqrt{|a|^2 - c}$.Calculating $|a|^2 = |i + 2j - k|^2 = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6$.Substituting the values,the radius is $\sqrt{6 - (-10)} = \sqrt{6 + 10} = \sqrt{16} = 4$.Thus,the equation represents a sphere of radius $4$.

The equation $|r|^2 - r \cdot (2i + 4j - 2k) - 10 = 0$ represents a

Similar Questions

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

$A$ 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

What is the center of the sphere passing through the four points $(0, 0, 0), (0, 2, 0), (1, 0, 0),$ and $(0, 0, 4)$?

$A$ 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:

The radius of the sphere $x^2+y^2+z^2=12x+4y+3z$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module