the radius of the circle in which the sphere x 2 y 2 z 2 2x

The radius of the circle in which the sphere $${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane $$x + 2y + 2z + 7 = 0$$ is :

A. $$4$$

B. $$1$$

C. $$2$$

D. $$3$$

Answer : $$3$$

Solution :
Three Dimensional Geometry mcq solution image

Centre of sphere $$ = \left( { - 1,\,1,\,2} \right)$$
Radius of sphere $$\sqrt {1 + 1 + 4 + 19} = 5$$
Perpendicular distance from centre to the plane
$$\eqalign{ & OC = d = \left| {\frac{{ - 1 + 2 + 4 + 7}}{{\sqrt {1 + 4 + 4} }}} \right| = \frac{{12}}{3} = 4 \cr & A{C^2} = A{O^2} - O{C^2} \cr & A{C^2} = {5^2} - {4^2} = 9 \cr & AC = 3 \cr} $$

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 :

A. $$7$$

B. $$ - 7$$

C. no real value

D. $$4$$

View Answer

Releted Question 2

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. $$\frac{3}{2}$$

B. $$\frac{9}{2}$$

C. $$ - \frac{2}{9}$$

D. $$ - \frac{3}{2}$$

View Answer

Releted Question 3

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 :

A. $$0$$

B. $$1$$

C. $$\sqrt 2 $$

D. $$2\sqrt 2 $$

View Answer

Releted Question 4

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 :

A. $$\frac{1}{4}$$

B. $$ - \frac{1}{4}$$

C. $$\frac{1}{8}$$

D. $$ - \frac{1}{8}$$

View Answer

The radius of the circle in which the sphere $${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane $$x + 2y + 2z + 7 = 0$$ is :

Releted MCQ Question on
Geometry >> Three Dimensional Geometry

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 :

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Three Dimensional Geometry

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The radius of the circle in which the sphere $${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane $$x + 2y + 2z + 7 = 0$$ is :

Releted MCQ Question on Geometry >> Three Dimensional Geometry

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 :

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