IIT-JEE 2010 Paper 2 Offline | 3D Geometry Question 31 | Mathematics

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

-1

If the distance of the point $$P(1, -2, 1)$$ from the plane $$x+2y-2z$$$$\, = \alpha ,$$ where $$\alpha > 0,$$ is $$5,$$ then the foot of the perpendicular from $$P$$ to the planes is

$$\left( {{8 \over 3},{4 \over 3}, - {7 \over 3}} \right)$$

$$\left( {{4 \over 3},-{4 \over 3}, {1 \over 3}} \right)$$

$$\left( {{1 \over 3},{2 \over 3}, {10 \over 3}} \right)$$

$$\left( {{2 \over 3},-{1 \over 3}, {5 \over 3}} \right)$$

IIT-JEE 2009 Paper 2 Offline

MCQ (Single Correct Answer)

-1

A line with positive direction cosines passes through the point P(2, $$-$$1, 2) and makes equal angles with the coordinate axes. The line meets the plane $$2x + y + z = 9$$ at point Q. The length of the line segment PQ equals

$$1$$

$${\sqrt 2 }$$

$${\sqrt 3 }$$

$$2$$

IIT-JEE 2009 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$P(3,2,6)$$ be a point in space and $$Q$$ be a point on the line $$$\widehat r = \left( {\widehat i - \widehat j + 2\widehat k} \right) + \mu \left( { - 3\widehat i + \widehat j + 5\widehat k} \right)$$$

Then the value of $$\mu $$ for which the vector $${\overrightarrow {PQ} }$$ is parallel to the plane $$x - 4y + 3z = 1$$ is :

$${1 \over 4}$$

$$-{1 \over 4}$$

$${1 \over 8}$$

$$-{1 \over 8}$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The distance of the point $$(1, 1, 1)$$ from the plane passing through the point $$(-1, -2, -1)$$ and whose normal is perpendicular to both the lines $${L_1}$$ and $${L_2}$$ is :

$${2 \over {\sqrt {75} }}$$

$${7 \over {\sqrt {75} }}$$

$${13 \over {\sqrt {75} }}$$

$${23 \over {\sqrt {75} }}$$

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