JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 14)
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 :
$$\sqrt {{A^2} + {B^2} + \sqrt 2 AB} $$
$$\sqrt {{A^2} + {B^2}} $$
$$\sqrt {{A^2} + {B^2} - \sqrt 2 AB} $$
$$\sqrt {{A^2} + {B^2} + 2AB} $$
Explanation
Given, $$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A \times \overrightarrow B } \right|$$ ..... (i)
Also, we know that
$$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ cos$$\theta$$ .... (ii)
and $$\overrightarrow A \times \overrightarrow B $$ = $$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ sin$$\theta$$ ..... (iii)
From Eqs. (i), (ii) and (iii), we get
$$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ cos$$\theta$$ = $$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ sin$$\theta$$
$$ \Rightarrow \cos \theta = \sin \theta \Rightarrow {{\sin \theta } \over {\cos \theta }} = 1$$
$$ \Rightarrow \tan \theta = 1$$
$$ \Rightarrow \tan \theta = \tan 45^\circ \Rightarrow \theta = 45^\circ $$
$$\therefore$$ $$\left| {\overrightarrow A - \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta } $$
$$ = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos (45^\circ )} $$
$$ = \sqrt {{A^2} + {B^2} - \sqrt 2 AB} $$
Also, we know that
$$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ cos$$\theta$$ .... (ii)
and $$\overrightarrow A \times \overrightarrow B $$ = $$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ sin$$\theta$$ ..... (iii)
From Eqs. (i), (ii) and (iii), we get
$$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ cos$$\theta$$ = $$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$$ sin$$\theta$$
$$ \Rightarrow \cos \theta = \sin \theta \Rightarrow {{\sin \theta } \over {\cos \theta }} = 1$$
$$ \Rightarrow \tan \theta = 1$$
$$ \Rightarrow \tan \theta = \tan 45^\circ \Rightarrow \theta = 45^\circ $$
$$\therefore$$ $$\left| {\overrightarrow A - \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta } $$
$$ = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos (45^\circ )} $$
$$ = \sqrt {{A^2} + {B^2} - \sqrt 2 AB} $$
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