In a quadrilateral ABCD, AC→ is the bisector of the (AB→ ^AD→) which is 2π3, 15|AC→|=3|AB→|=5|AD→|, if cos2(BA→ ^CD→)=pq (where p and q are co-prime and (a→, b→)denotes the angle between a→ and b→ ) then p+q is equal to.

15|AC¯|=3|AB¯|=5|AD¯| Let |AC¯|=λ=0,|AB¯|=5λ,|AD¯|=3λcos⁡BA¯∧CD¯=BA¯⋅CD¯|BA¯||CD¯|=−b¯⋅(d¯−c¯)|b¯||d¯−c¯| Numerator=|b¯||c¯|cos⁡π3−|b¯||d¯|cos⁡2π3=(5λ)(λ)12+(5λ)(3λ)12=5λ2+15λ22=10λ2 Now |d¯−c¯|2=d¯+c¯2−2c¯⋅d¯9λ2+λ2−2(λ)(3λ)1/2=10λ2−3λ2=7λ2∴|d¯−c¯|=7λ.cos⁡BA¯∧CD¯=10λ257λ2=27

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In a quadrilateral ABCD, $\vec{A C}$ is the bisector of the $(\vec{A B} \hat{} \vec{A D})$ which is $\frac{2 π}{3},$ $15 | \vec{A C} | = 3 | \vec{A B} | = 5 | \vec{A D} |,$ if $\cos^{2} (\vec{B A} \hat{} \vec{C D}) = \frac{p}{q}$ (where p and q are co-prime and ( $\vec{a}$ , $\vec{b}$ )denotes the angle between $\vec{a}$ and $\vec{b}$ ) then $p + q$ is equal to.

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Detailed Solution

$15 | \bar{A C} | = 3 | \bar{A B} | = 5 | \bar{A D} | L e t \begin{aligned} | \bar{AC} | = λ = 0, | \bar{AB} | = 5 λ, | \bar{AD} | = 3 λ \\ \cos ({\bar{BA}}^{\land} \bar{CD}) = \frac{\bar{BA} \cdot \bar{CD}}{| \bar{BA} | | \bar{CD} |} = \frac{- \bar{b} \cdot (\bar{d} - \bar{c})}{| \bar{b} | | \bar{d} - \bar{c} |} \end{aligned}$

$\begin{aligned} N u m e r a t o r = | \bar{b} | | \bar{c} | \cos \frac{π}{3} - | \bar{b} | | \bar{d} | \cos \frac{2 π}{3} \\ = (5 λ) (λ) \frac{1}{2} + (5 λ) (3 λ) \frac{1}{2} = \frac{5 λ^{2} + 15 λ^{2}}{2} = 10 λ^{2} \end{aligned}$

$\begin{aligned} Now | \bar{d} - \bar{c} |^{2} = \bar{d} + {\bar{c}}^{2} - 2 \bar{c} \cdot \bar{d} \\ 9 λ^{2} + λ^{2} - 2 (λ) (3 λ) 1 / 2 = 10 λ^{2} - 3 λ^{2} = 7 λ^{2} \\ ∴ | \bar{d} - \bar{c} | = \sqrt{7} λ . \\ \cos \bar{BA} \land \bar{CD} = \frac{10 λ^{2}}{5 \sqrt{7} λ^{2}} = \frac{2}{\sqrt{7}} \end{aligned}$