IIT-JEE 2010 Paper 2 Offline | Vector Algebra Question 29 | Mathematics

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Two adjacent sides of a parallelogram $$ABCD$$ are given by
$$\overrightarrow {AB} = 2\widehat i + 10\widehat j + 11\widehat k$$ and $$\,\overrightarrow {AD} = -\widehat i + 2\widehat j + 2\widehat k$$
The side $$AD$$ is rotated by an acute angle $$\alpha $$ in the plane of the parallelogram so that $$AD$$ becomes $$AD'.$$ If $$AD'$$ makes a right angle with the side $$AB,$$ then the cosine of the angle $$\alpha $$ is given by

$${{8 \over 9}}$$

$${{{\sqrt {17} } \over 9}}$$

$${{1 \over 9}}$$

$${{{4\sqrt 5 } \over 9}}$$

IIT-JEE 2009 Paper 1 Offline

MCQ (Single Correct Answer)

-1

If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $$\overrightarrow d $$ are unit vectors such that $$(\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d ) = 1$$ and $$\overrightarrow a \,.\,\overrightarrow c = {1 \over 2}$$, then

$$\overrightarrow a \,,\,\overrightarrow b ,\overrightarrow c $$ are non-coplanar

$$\overrightarrow b \,,\,\overrightarrow c ,\overrightarrow d $$ are non-coplanar

$$\overrightarrow b \,,\overrightarrow d $$ are non-parallel

$$\overrightarrow a ,\overrightarrow d $$ parallel and $$\overrightarrow b ,\overrightarrow c $$ are parallel

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

$$0$$

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

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

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

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