IIT-JEE 2008 Paper 2 Offline | Vector Algebra and 3D Geometry Question 63 | Mathematics

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 unit vector perpendicular to both $${L_1}$$ and $${L_2}$$ is :

$${{ - \widehat i + 7\widehat j + 7\widehat k} \over {\sqrt {99} }}$$

$${{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$

$${{ - \widehat i + 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$

$${{7\widehat i - 7\widehat j - \widehat k} \over {\sqrt {99} }}$$

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 :

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

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

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

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

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 shortest distance between $${L_1}$$ and $${L_2}$$ is :

$$0$$

$${17 \over {\sqrt 3 }}$$

$${41 \over {5\sqrt 3 }}$$

$${17 \over {5\sqrt 3 }}$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $$\overrightarrow a \,,\,\overrightarrow b ,\overrightarrow c $$ such that $$\widehat a\,.\,\widehat b = \widehat b\,.\,\widehat c = \widehat c\,.\,\widehat a = {1 \over 2}.$$ Then, the volume of the parallelopiped is :

$${1 \over {\sqrt 2 }}$$

$${1 \over {2\sqrt 2 }}$$

$${{\sqrt 3 } \over 2}$$

$${1 \over {\sqrt 3 }}$$

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