Let two non-collinear vectors a→ and b→ inclined at an angle 2π3 be such that |a→|=4 and |b→|=3 . A point P moves so that at any time t the position vector OP¯ (Where O is the origin) is given as OP¯ = (et+e−t)a¯+(et−e−t)b.¯ If the least distance of P from origin is 2 m−n units, Where m,n∈I, then the value of (4m−14n+2) is

we have (OP¯)2=(et+e−t)2(a¯)2+(et−e−t)2(b¯)2+2(et+e−t)(et−e−t)(a¯ . b¯)⇒(OP¯)2=16(et+e−t)+9(et−e−t)+2(e2t−e−2t).3.4.(−12)=13e2t+37e−2t+14≥213×37+14⇒|OP¯|≥2481+14⇒|OP¯|≥2481−(−7)By Comparing with the given question a=481, b=−7∴(4a−14b+2)=4(481)−14(−7)+2 =2024

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Let two non-collinear vectors $\vec{a} a n d \vec{b}$ inclined at an angle $\frac{2 π}{3}$ be such that $| \vec{a} | = 4$ and $| \vec{b} | = 3$ . A point P moves so that at any time t the position vector $\bar{O P}$ (Where O is the origin) is given as $\bar{O P}$ = $(e^{t} + e^{- t}) \bar{a} + (e^{t} - e^{- t}) \bar{b .}$ If the least distance of P from origin is $\sqrt{2} \sqrt{\sqrt{m} - n}$ units, Where $m, n \in I,$ then the value of $(4 m - 14 n + 2)$ is

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answer is 2024.

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Detailed Solution

we have ${(\bar{O P})}^{2} = {(e^{t} + e^{- t})}^{2} {(\bar{a})}^{2} + {(e^{t} - e^{- t})}^{2} {(\bar{b})}^{2} + 2 (e^{t} + e^{- t}) (e^{t} - e^{- t}) (\bar{a} . \bar{b})$

$\begin{array}{l} \Rightarrow {(\bar{O P})}^{2} = 16 (e^{t} + e^{- t}) + 9 (e^{t} - e^{- t}) + 2 (e^{2 t} - e^{- 2 t}) .3.4. (- \frac{1}{2}) \\ = 13 e^{2 t} + 37 e^{- 2 t} + 14 \geq 2 \sqrt{13 \times 37} + 14 \\ \Rightarrow | \bar{O P} | \geq \sqrt{2 \sqrt{481} + 14} \\ \Rightarrow | \bar{O P} | \geq \sqrt{2} \sqrt{\sqrt{481} - (- 7)} \\ B y C o m p a r i n g w i t h t h e g i v e n q u e s t i o n a = 481, b = - 7 \\ ∴ (4 a - 14 b + 2) = 4 (481) - 14 (- 7) + 2 = 2024 \end{array}$