Question 74: Motion in 2D & Vectors

Question

If $$\overrightarrow A $$ and $$\overrightarrow B $$ are two vectors satisfying the relation $$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A  	imes \overrightarrow B } ight|$$. Then the value of $$\left| {\overrightarrow A  - \overrightarrow B } ight|$$ will be :

Accepted Answer

Given, $$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A  	imes \overrightarrow B } ight|$$ ..... (i)
Also, we know that
$$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A } ight|\left| {\overrightarrow B } ight|$$ cos$$	heta$$ .... (ii)
and $$\overrightarrow A  	imes \overrightarrow B $$ = $$\left| {\overrightarrow A } ight|\left| {\overrightarrow B } ight|$$ sin$$	heta$$ ..... (iii)
From Eqs. (i), (ii) and (iii), we get
$$\left| {\overrightarrow A } ight|\left| {\overrightarrow B } ight|$$ cos$$	heta$$ = $$\left| {\overrightarrow A } ight|\left| {\overrightarrow B } ight|$$ sin$$	heta$$
$$ \Rightarrow \cos 	heta  = \sin 	heta  \Rightarrow {{\sin 	heta } \over {\cos 	heta }} = 1$$
$$ \Rightarrow 	an 	heta  = 1$$
$$ \Rightarrow 	an 	heta  = 	an 45^\circ  \Rightarrow 	heta  = 45^\circ $$
$$	herefore$$ $$\left| {\overrightarrow A  - \overrightarrow B } ight| = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } ight|\left| {\overrightarrow B } ight|\cos 	heta } $$
$$ = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } ight|\left| {\overrightarrow B } ight|\cos (45^\circ )} $$
$$ = \sqrt {{A^2} + {B^2} - \sqrt 2 AB} $$

Answer

$$\sqrt {{A^2} + {B^2} + \sqrt 2 AB} $$

Answer

$$\sqrt {{A^2} + {B^2}} $$

Answer

$$\sqrt {{A^2} + {B^2} - \sqrt 2 AB} $$

Answer

$$\sqrt {{A^2} + {B^2} + 2AB} $$

Motion in 2D & Vectors

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 :

Choose the correct answer: