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1
JEE Main 2021 (Online) 20th July Morning Shift
+4
-1
If $$\overrightarrow A$$ and $$\overrightarrow B$$ are two vectors satisfying the relation $$\overrightarrow A$$ . $$\overrightarrow B$$ = $$\left| {\overrightarrow A \times \overrightarrow B } \right|$$. Then the value of $$\left| {\overrightarrow A - \overrightarrow B } \right|$$ will be :
A
$$\sqrt {{A^2} + {B^2} + \sqrt 2 AB}$$
B
$$\sqrt {{A^2} + {B^2}}$$
C
$$\sqrt {{A^2} + {B^2} - \sqrt 2 AB}$$
D
$$\sqrt {{A^2} + {B^2} + 2AB}$$
2
JEE Main 2021 (Online) 25th February Morning Slot
+4
-1
In an octagon ABCDEFGH of equal side, what is the sum of

$$\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} + \overrightarrow {AG} + \overrightarrow {AH}$$,

if, $$\overrightarrow {AO} = 2\widehat i + 3\widehat j - 4\widehat k$$ A
$$- 16\widehat i - 24\widehat j + 32\widehat k$$
B
$$16\widehat i + 24\widehat j - 32\widehat k$$
C
$$16\widehat i + 24\widehat j + 32\widehat k$$
D
$$16\widehat i - 24\widehat j + 32\widehat k$$
3
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
Let $$\left| {\mathop {{A_1}}\limits^ \to } \right| = 3$$, $$\left| {\mathop {{A_2}}\limits^ \to } \right| = 5$$ and $$\left| {\mathop {{A_1}}\limits^ \to + \mathop {{A_2}}\limits^ \to } \right| = 5$$. The value of $$\left( {2\mathop {{A_1}}\limits^ \to + 3\mathop {{A_2}}\limits^ \to } \right)\left( {3\mathop {{A_1}}\limits^ \to - \mathop {2{A_2}}\limits^ \to } \right)$$ is :-
A
–118.5
B
–112.5
C
–99.5
D
–106.5
4
JEE Main 2019 (Online) 10th January Evening Slot
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 -
$${\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$$
$${\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$$
$${\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$$
$${\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$$