For $p>0$, a vector $\vec{v}_{2}=2 \hat{i}+(p+1) \hat{j}$ is obtained by rotating the vector $\vec{v}_{1}=\sqrt{3} p \hat{i}+\hat{j}$ by an angle $\theta$ about origin in counter clockwise direction. If $\tan \theta=\frac{(\alpha \sqrt{3}-2)}{(4 \sqrt{3}+3)}$, then the value of $\alpha$ is equal to

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Vector Algebra

For

p > 0

, a vector

v_{2} = 2 \hat{i} + (p + 1) \hat{j}

is obtained by rotating the vector

v_{1} = 3 p \hat{i} + \hat{j}

by an angle

θ

about origin in counter clockwise direction. If

tan θ = \frac{( α 3 - 2 )}{( 4 3 + 3 )}

, then the value of

α

is equal to

Answer: 6

When a vector is rotated by angle

θ

counterclockwise, the magnitudes remain equal:

∣ v_{1} ∣ = ∣ v_{2} ∣

.

So

3 p^{2} + 1 = 4 + (p + 1)^{2}

Squaring:

3 p^{2} + 1 = 4 + p^{2} + 2 p + 1

2 p^{2} - 2 p - 4 = 0 \Rightarrow p^{2} - p - 2 = 0 \Rightarrow (p - 2) (p + 1) = 0

Since

p > 0

, we have

p = 2

.

Thus

v_{1} = 23 \hat{i} + \hat{j}

and

v_{2} = 2 \hat{i} + 3 \hat{j}

.

The angle of

v_{1}

with x-axis:

tan θ_{1} = \frac{1}{2 3}

The angle of

v_{2}

with x-axis:

tan θ_{2} = \frac{3}{2}

Rotation angle:

θ = θ_{2} - θ_{1}

Using

tan (θ_{2} - θ_{1}) = \frac{t a n θ _{2} - t a n θ _{1}}{1 + t a n θ _{1} t a n θ _{2}} = \frac{\frac{3}{2} - \frac{1}{2 3}}{1 + \frac{3}{4 3}} = \frac{6 3 - 2}{4 3 + 3}

Comparing with the given form,

α = 6

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer