Question 59: Motion in 2D & Vectors

Question

The angle between vector $\left( {\overrightarrow A } \right)$ and $\left( {\overrightarrow A  - \overrightarrow B } \right)$ is :

Accepted Answer

Angle between ${\overrightarrow A }$ and ${\overrightarrow B }$, $	heta$ = 60$^\circ$
Angle between ${\overrightarrow A }$ and -${\overrightarrow B }$, $	heta$ = 120$^\circ$
If angle between ${\overrightarrow A }$ and ${\overrightarrow A }$ $-$ ${\overrightarrow B }$ is $\alpha $
then $	an \alpha  = $${{\left| { - \overrightarrow B } ight|\sin 	heta } \over {\overrightarrow A  + \left| { - \overrightarrow B } ight|\cos 	heta }}$
= ${{B\sin 120^\circ } \over {A + B\cos 120^\circ }}$
=${{B{{\sqrt 3 } \over 2}} \over {A - {B \over 2}}}$
$ \Rightarrow $ $	an \alpha  = {{\sqrt 3 B} \over {2A - B}}$

Answer

${	an ^{ - 1}}\left( {{{ - {B \over 2}} \over {A - B{{\sqrt 3 } \over 2}}}} ight)$

Answer

${	an ^{ - 1}}\left( {{A \over {0.7B}}} ight)$

Answer

${	an ^{ - 1}}\left( {{{\sqrt 3 B} \over {2A - B}}} ight)$

Answer

${	an ^{ - 1}}\left( {{{B\cos 	heta } \over {A - B\sin 	heta }}} ight)$

Motion in 2D & Vectors

The angle between vector \left( {\overrightarrow A } \right) and \left( {\overrightarrow A - \overrightarrow B } \right) is :

Choose the correct answer: