For p-0- a vector V2-2-p-1- is obtained by rotating the vector V1-3p- by an angle - about origin in counter clockwise direction- If tan- - -3-2 4-3-3 - then the value of - is equal to

∵ nbsp; cosθ =V_1.V_2|V_1|.|V_2| and |V_1|=|V_2| ⇒ nbsp; cosθ =2√(3p)+p+1|V_1|^2 and 4+(p+1)^2=3p^2+1 ⇒ p=2 ⇒cosθ =4√(3)+313⇒tanθ =6√(3) 24√(3)+3 ...

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Standard XII

Physics

Summary

For p>0, a ve...

Question

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

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Solution

$∵ cos θ = \frac{_{1} ._{2}}{|_{1} | . |_{2} |} and |_{1} | = |_{2} |$
$\Rightarrow cos θ = \frac{2 \sqrt{3 p} + p + 1}{|_{1} |^{2}} and 4 + (p + 1)^{2} = 3 p^{2} + 1$
$\Rightarrow p = 2$
$\Rightarrow cos θ = \frac{4 \sqrt{3} + 3}{13} \Rightarrow tan θ = \frac{6 \sqrt{3} - 2}{4 \sqrt{3} + 3}$

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For p>0, a vector →V2=2^i+(p+1)^j is obtained by rotating the vector →V1=√3p^i+^j by an angle θ about origin in counter clockwise direction. If tanθ=(α√3−2)(4√3+3), then the value of α is equal to

∵cosθ=→V1.→V2|→V1|.|→V2| and |→V1|=|→V2| ⇒cosθ=2√3p+p+1|→V1|2 and 4+(p+1)2=3p2+1 ⇒p=2 ⇒cosθ=4√3+313⇒tanθ=6√3−24√3+3

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

$∵ cos θ = \frac{_{1} ._{2}}{|_{1} | . |_{2} |} and |_{1} | = |_{2} |$
$\Rightarrow cos θ = \frac{2 \sqrt{3 p} + p + 1}{|_{1} |^{2}} and 4 + (p + 1)^{2} = 3 p^{2} + 1$
$\Rightarrow p = 2$
$\Rightarrow cos θ = \frac{4 \sqrt{3} + 3}{13} \Rightarrow tan θ = \frac{6 \sqrt{3} - 2}{4 \sqrt{3} + 3}$