JEE Main 2018 (Online) 15th April Evening Slot | 3D Geometry Question 337 | Mathematics

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

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Out of Syllabus

A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point :

($$-$$ 3, 2, 1)

(3, 2, 1)

($$-$$ 1, 2, 3)

(1, 2, $$-$$ 3)

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

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Out of Syllabus

A variable plane passes through a fixed point (3,2,1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz -plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is :

$${x \over 3} + {y \over 2} + {z \over 1} = 1$$

x + y + z = 6

$${1 \over x} + {1 \over y} + {1 \over z} = {{11} \over 6}$$

$${3 \over x} + {2 \over y} + {1 \over z} = 1$$

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

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Out of Syllabus

An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z $$-$$ 1 = 0 and 5x + 8y + 2z + 14 =0, is :

$${\sin ^{ - 1}}\left( {\sqrt {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {17}$}}} } \right)$$

$${\cos ^{ - 1}}\left( {\sqrt {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {17}$}}} } \right)$$

$${\cos ^{ - 1}}\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {17}$}}} \right)$$

$${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {17}$}}} \right)$$

JEE Main 2017 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

Out of Syllabus

If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of $$\Delta $$ABC is :

$${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 3$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = {1 \over 9}$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 9$$

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