AIEEE 2004 | 3D Geometry Question 354 | Mathematics

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

$$2x-y-z=1$$

$$x-2y-z=1$$

$$x-y-2z=1$$

$$x-y-z=1$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

Out of Syllabus

The shortest distance from the plane $$12x+4y+3z=327$$ to the sphere

$${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is

$$39$$

$$26$$

$$11{4 \over {13}}$$

$$13$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

The two lines $$x=ay+b,z=cy+d$$ and $$x = a'y + b',z = c'y + d'$$ will be perpendicular, if and only if :

$$aa' + cc' + 1 = 0$$

$$aa' + bb'cc' + 1 = 0$$

$$aa' + bb'cc' = 0$$

$$\left( {a + a'} \right)\left( {b + b'} \right) + \left( {c + c'} \right) = 0$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

The lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar if :

$$k=3$$ or $$-2$$

$$k=0$$ or $$-1$$

$$k=1$$ or $$-1$$

$$k=0$$ or $$-3$$

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