Question

(b)
The angle between vectors A and B is 600. What is the ratio of A B and | AXB|?​

Answer 1

Appropriate Question:-

The angle between vectors $\sf \overrightarrow{A}$ and $\sf\overrightarrow{B}$ is 60°. What is the ratio of

$\sf\overrightarrow{A}.\overrightarrow{B}$ and | $\sf\overrightarrow{A}\times\overrightarrow{B}$| ?​

Solution:-

We are provided with the angle between two vectors and we are asked to find the ratio of their dot product and cross product.

Calculate them individually and then find their ratio.

Dot Product:-

$\quad\mapsto\sf\overrightarrow{A}.\overrightarrow{B} = A B \cos\theta$

$\quad\mapsto\sf\overrightarrow{A}.\overrightarrow{B} = A B \cos 60$

$\quad\mapsto\sf\overrightarrow{A}.\overrightarrow{B} = \dfrac{AB}{2}$

Cross Product:-

$\quad\mapsto\sf\overrightarrow{A}\times\overrightarrow{B} = A B \sin\theta$

$\quad\mapsto\sf\overrightarrow{A}\times\overrightarrow{B} = A B \sin60$

$\quad\mapsto\sf\overrightarrow{A}\times\overrightarrow{B} = \dfrac{AB\sqrt{3}}{2}$

Now find the ratio:-

$\\\quad:\implies\sf \dfrac{\overrightarrow{A}.\overrightarrow{B}}{\overrightarrow{A}\times\overrightarrow{B}} =\dfrac{AB}{2} \times \dfrac{2}{AB\sqrt{3} }$

$\\\quad:\implies\boxed{\sf \dfrac{\overrightarrow{A}.\overrightarrow{B}}{\overrightarrow{A}\times\overrightarrow{B}} = \dfrac{1}{\sqrt{3} }}\bigstar$

∴ the required ratio is 1:√3

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