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3-1.Vectors
easy
For any two vectors $\overrightarrow A $ and $\overrightarrow B $, if $\overrightarrow A \,.\,\overrightarrow B = \,\,|\overrightarrow A \times \overrightarrow B |,$ the magnitude of $\overrightarrow C = \overrightarrow A + \overrightarrow B $ is equal to
A
$\sqrt {{A^2} + {B^2}} $
B
$A + B$
C
$\sqrt {{A^2} + {B^2} + \frac{{AB}}{{\sqrt 2 }}} $
D
$\sqrt {{A^2} + {B^2} + \sqrt 2 \times AB} $
Solution
(d) $AB\cos \theta = AB\sin \theta $==>$\tan \theta = 1$? $\theta = 45^\circ $
$\therefore |\bar C|\, = \sqrt {{A^2} + {B^2} + 2AB\cos 45^\circ } $$ = \sqrt {{A^2} + {B^2} + \sqrt 2 AB} $
Standard 11
Physics
Similar Questions
Vector $A$ is pointing eastwards and vector $B$ northwards. Then, match the following two columns.
| Colum $I$ | Colum $II$ |
| $(A)$ $(A+B)$ | $(p)$ North-east |
| $(B)$ $(A-B)$ | $(q)$ Vertically upwards |
| $(C)$ $(A \times B)$ | $(r)$ Vertically downwards |
| $(D)$ $(A \times B) \times(A \times B)$ | $(s)$ None |
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