Exams Joint Entrance Examination Mathematics Vector Algebra Question 39 Question 39 of 379 Hard MCQ
Let the arc A C A C A C of a circle subtend a right angle at the centre O O O . If the point B B B on the arc A C A C A C , divides the arc A C A C A C
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c → = α a → + β b → … . ( 1 ) a → . c → = α a → ⋅ a → + β b → ⋅ a → 0 = α + β cos 15 ∘ … . ( 2 ) ( 1 ) ⇒ b → ⋅ c → = α a → ⋅ b → ← + β b → ⋅ b → ← ⇒ cos 75 ∘ = α cos 15 ∘ + β … . ( 3 ) ( 2 ) ( 3 ) ⇒ cos 75 ∘ = − β cos 2 15 ∘ + β \begin{gathered} \overset{\to }{c}=\alpha \overset{\to }{a}+\beta \overset{\to }{b}\ldots .(1) \\ \overset{\to }{a}.\overset{\to }{c}=\alpha \overset{\to }{a}\cdot \overset{\to }{a}+\beta \overset{\to }{b}\cdot \overset{\to }{a} \\ 0=\alpha +\beta \cos {15}^{\circ }\ldots .(2) \\ (1)\Rightarrow \overset{\to }{b}\cdot \overset{\to }{c}=\alpha \overset{\to }{a}\cdot \overset{\to \leftarrow }{b}+\beta \overset{\to }{b}\cdot \overset{\to \leftarrow }{b} \\ \Rightarrow \cos {75}^{\circ }=\alpha \cos {15}^{\circ }+\beta \ldots .(3) \\ (2)(3)\Rightarrow \cos {75}^{\circ }=-\beta {\cos }^{2}{15}^{\circ }+\beta \end{gathered} c → = α a → + β b → … . ( 1 ) a → . c → = α a → ⋅ a → + β b → ⋅ a → 0 = α + β cos 15 ∘ … . ( 2 ) ( 1 ) ⇒ b → ⋅ c → = α a → ⋅ b →← + β b → ⋅ b →← ⇒ cos 75 ∘ = α cos 15 ∘ + β … . ( 3 ) ( 2 ) ( 3 ) ⇒ cos 75 ∘ = − β cos 2 15 ∘ + β
β = cos 75 ∘ sin 2 15 ∘ = 1 sin 15 ∘ = 2 2 3 − 1 ( 2 ) ⇒ α = − cos 15 ∘ sin 15 ∘ = − ( 3 + 1 ) ( 3 − 1 ) ∴ c → = − ( 3 + 1 ) ( 3 − 1 ) a → + ( 2 2 3 − 1 ) b → \begin{gathered} \beta =\frac{\cos {75}^{\circ }}{{\sin }^{2}{15}^{\circ }}=\frac{1}{\sin {15}^{\circ }}=\frac{2\sqrt{2}}{\sqrt{3}-1} \\ (2)\Rightarrow \alpha =\frac{-\cos {15}^{\circ }}{\sin {15}^{\circ }}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)} \\ \therefore \overset{\to }{c}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}\overset{\to }{a}+(\frac{2\sqrt{2}}{\sqrt{3}-1})\overset{\to }{b} \end{gathered} β = sin 2 15 ∘ cos 75 ∘ = sin 15 ∘ 1 = 3 − 1 2 2 ( 2 ) ⇒ α = sin 15 ∘ − cos 15 ∘ = ( 3 − 1 ) − ( 3 + 1 ) ∴ c → = ( 3 − 1 ) − ( 3 + 1 ) a → + ( 3 − 1 2 2 ) b →
Now
α + 2 ( 3 − 1 ) β = − ( 3 + 1 ) ( 3 − 1 ) + 2 ( 3 − 1 ) ⋅ 2 2 3 − 1 = − ( 3 + 1 ) 2 2 + 4 = − 3 − 1 − 2 3 + 8 2 = 2 − 3 \begin{gathered} \alpha +\sqrt{2}(\sqrt{3}-1)\beta =\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{gathered} α + 2 ( 3 − 1 ) β = ( 3 − 1 ) − ( 3 + 1 ) + 3 − 1 2 ( 3 − 1 ) ⋅ 2 2 = 2 − ( 3 + 1 ) 2 + 4 = 2 − 3 − 1 − 2 3 + 8 = 2 − 3 Answer the question to unlock