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 sphere and the plane :
A.
$$2x-y-z=1$$
B.
$$x-2y-z=1$$
C.
$$x-y-2z=1$$
D.
$$x-y-z=1$$
Answer :
$$2x-y-z=1$$
Solution :
The equations of spheres are $${S_1} : {x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$$ and $${S_2} : {x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$$
Their plane of intersection is
$$\eqalign{
& {S_1} - {S_2} = 0 \cr
& \Rightarrow 10x - 5y - 5z - 5 = 0 \cr
& \Rightarrow 2x - y - z = 1 \cr} $$
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :