JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 17)
For p > 0, a vector $${\overrightarrow v _2} = 2\widehat i + (p + 1)\widehat j$$ is obtained by rotating the vector $${\overrightarrow v _1} = \sqrt 3 p\widehat i + \widehat j$$ by an angle $$\theta$$ about origin in counter clockwise direction. If $$\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\left( {4\sqrt 3 + 3} \right)}}$$, then the value of $$\alpha$$ is equal to _____________.
Answer
6
Explanation
_20th_July_Evening_Shift_en_17_2.png)
$$\left| {\overrightarrow {{V_1}} } \right| = \left| {\overrightarrow {{V_2}} } \right|$$
$$3{P^2} + 1 = 4 + {(P + 1)^2}$$
$$2{P^2} - 2P - 4 = 0 \Rightarrow {P^2} - P - 2 = 0$$
$$P = 2, - 1$$ (rejected)
$$\cos \theta = {{\overrightarrow {{V_1}} .\overrightarrow {{V_2}} } \over {\left| {\overrightarrow {{V_1}} } \right|\left| {\overrightarrow {{V_2}} } \right|}} = {{2\sqrt 3 P + (P + 1)} \over {\sqrt {{{(P + 1)}^2} + 4} \sqrt {3{P^2} + 1} }}$$
$$\cos \theta = {{4\sqrt 3 + 3} \over {\sqrt {13} \sqrt {13} }} = {{4\sqrt 3 + 3} \over {13}}$$
$$\tan \theta = {{\sqrt {112 - 24\sqrt 3 } } \over {4\sqrt 3 + 3}} = {{6\sqrt 3 - 2} \over {4\sqrt 3 + 3}} = {{\alpha \sqrt 3 - 2} \over {4\sqrt 3 + 3}}$$
$$ \Rightarrow \alpha = 6$$
Comments (0)
