Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that length of arcAB length of arcBC=15, and OC→=αOA→+βOB→, thenα=2(3-1)β is equal to

c→=αa→+βb→….(1)a→⋅c→=αa→⋅a→+βb→⋅a→0=α+βcos⁡15∘….(2)(1)⇒b→⋅c→=αa→⋅b→+βb→⋅b→⇒cos⁡75∘=αcos⁡15∘+β…β=cos⁡75∘sin2⁡15∘=1sin⁡15∘=223−1 (2) ⇒α=−cos⁡15∘sin⁡15∘=−(3+1)(3−1)∴c→=−(3+1)(3−1)a→+223−1b→α+2(3−1)β=−(3+1)(3−1)+2(3−1)⋅223−1=−(3+1)22+4=−3−1−23+82=2−3

Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that length of arcAB length of arcBC=15, and OC→=αOA→+βOB→, thenα=2(3-1)β is equal to

c→=αa→+βb→….(1)a→⋅c→=αa→⋅a→+βb→⋅a→0=α+βcos⁡15∘….(2)(1)⇒b→⋅c→=αa→⋅b→+βb→⋅b→⇒cos⁡75∘=αcos⁡15∘+β…β=cos⁡75∘sin2⁡15∘=1sin⁡15∘=223−1 (2) ⇒α=−cos⁡15∘sin⁡15∘=−(3+1)(3−1)∴c→=−(3+1)(3−1)a→+223−1b→α+2(3−1)β=−(3+1)(3−1)+2(3−1)⋅223−1=−(3+1)22+4=−3−1−23+82=2−3

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Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that $\frac{length of arc AB}{length of arc BC} = \frac{1}{5},$ and $\vec{OC} = α \vec{OA} + β \vec{OB},$ then $α = \sqrt{2} (\sqrt{3} - 1) β$ is equal to

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2 $\sqrt{3}$

5 $\sqrt{3}$

2 + $\sqrt{3}$

2 - $\sqrt{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{c} = α \vec{a} + β \vec{b} \dots . (1) \\ \vec{a} \cdot \vec{c} = α \vec{a} \cdot \vec{a} + β \vec{b} \cdot \vec{a} \\ 0 = α + β \cos 15^{\circ} \dots . (2) \\ (1) \Rightarrow \vec{b} \cdot \vec{c} = α \vec{a} \cdot \vec{b} + β \vec{b} \cdot \vec{b} \\ \Rightarrow \cos 75^{\circ} = α \cos 15^{\circ} + β \dots \end{array}$

$β = \frac{\cos 75^{\circ}}{\sin^{2} 15^{\circ}} = \frac{1}{\sin 15^{\circ}} = \frac{2 \sqrt{2}}{\sqrt{3} - 1} \begin{aligned} (2) \Rightarrow α = \frac{- \cos 15^{\circ}}{\sin 15^{\circ}} = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)} \\ ∴ \vec{c} = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)} \vec{a} + (\frac{2 \sqrt{2}}{\sqrt{3} - 1}) \vec{b} \end{aligned}$

$\begin{array}{l} α + \sqrt{2} (\sqrt{3} - 1) β = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)} + \frac{\sqrt{2} (\sqrt{3} - 1) \cdot 2 \sqrt{2}}{\sqrt{3} - 1} \\ = \frac{- (\sqrt{3} + 1)^{2}}{2} + 4 \\ = \frac{- 3 - 1 - 2 \sqrt{3} + 8}{2} \\ = 2 - \sqrt{3} \end{array}$