JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Two vectors $$\overrightarrow A $$ and $$\overrightarrow B $$ have equal magnitudes. The magnitude of $$\left( {\overrightarrow A + \overrightarrow B } \right)$$ is 'n' times the magnitude of $$\left( {\overrightarrow A - \overrightarrow B } \right)$$ . The angle between $${\overrightarrow A }$$ and $${\overrightarrow B }$$ is -

$${\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$$

$${\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$$

$${\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$$

$${\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$$

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be -

JEE Main 2019 (Online) 10th January Morning Slot Physics - Vector Algebra Question 24 English

JEE Main 2019 (Online) 10th January Morning Slot Physics - Vector Algebra Question 24 English

$${1 \over 2}a\left( {\widehat k - \widehat i} \right)$$

$${1 \over 2}a\left( {\widehat j - \widehat i} \right)$$

$${1 \over 2}a\left( {\widehat j - \widehat k} \right)$$

$${1 \over 2}a\left( {\widehat i - \widehat k} \right)$$

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

Let $$\overrightarrow A $$ = $$\left( {\widehat i + \widehat j} \right)$$ and, $$\overrightarrow B = \left( {2\widehat i - \widehat j} \right).$$ The magnitude of a coplanar vector $$\overrightarrow C $$ such that $$\overrightarrow A .\overrightarrow C = \overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B ,$$ is given by :

$$\sqrt {{{10} \over 9}} $$

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

$$\sqrt {{{20} \over 9}} $$

$$\sqrt {{{9} \over 12}} $$

AIEEE 2004

MCQ (Single Correct Answer)

-1

If $$\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A $$, then the angle beetween A and B is

$${\pi \over 2}$$

$${\pi \over 3}$$

$$\pi $$

$${\pi \over 4}$$