Let a - b - c be three complex numbers whose modulus is 1 - If a - b cos- c sin-0- where -0- -2- thenA- b2-c2-0B- b2-c2-0C- a - be i -D- a-b e-i -

A+bcosα +c sinα = 0 ⇒ a = b cosα c sinα ⇒ |a|^2 = |b cosα +c sinα|^2 ⇒ 1 = (b cosα +c sinα)(b̅cosα + c̅sinα) ⇒ b b̅cos^2 α +c c̅sin^2 α nbsp; +12 (b c̅ +b̅ ...

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Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

Let a , b , c...

Question

Let $a, b, c$ be three complex numbers whose modulus is $1$ . If $a + b cos α + c sin α = 0$ , where $α \in (0, \frac{π}{2})$ , then

b^{2} + c^{2} = 0

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b^{2} - c^{2} = 0

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a = - b e^{i α}

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a = - b e^{- i α}

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Solution

The correct options are
A $b^{2} + c^{2} = 0$
C $a = - b e^{i α}$
D $a = - b e^{- i α}$
$a + b cos α + c sin α = 0 \Rightarrow a = - b cos α - c sin α \Rightarrow | a |^{2} = | b cos α + c sin α |^{2} \Rightarrow 1 = (b cos α + c sin α) (¯ b cos α + ¯ c sin α) \Rightarrow b ¯ b {cos}^{2} α + c ¯ c {sin}^{2} α + \frac{1}{2} (b ¯ c + ¯ b c) sin 2 α = 1 \Rightarrow \frac{1}{2} (b ¯ c + ¯ b c) sin 2 α = 0$
As $α \in (0, \frac{π}{2})$
$\Rightarrow (b ¯ c + ¯ b c) = 0 \Rightarrow \frac{b}{¯ b} = - \frac{c}{¯ c} \Rightarrow b^{2} = - c^{2} \Rightarrow b^{2} + c^{2} = 0 \Rightarrow c = \pm i b$

Using this, we get
$a = - (b cos α \pm i b sin α) = - b e^{\pm i α}$

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Similar questions

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a, b, c

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α \in (0, \frac{π}{2})

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and

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} + c^{2} & a b & a c a b & c^{2} + a^{2} & b c a c & b c & a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = K a^{2} b^{2} c^{2}

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a^{2} + b^{2} + c^{2} = 0

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Let a,b,c be three complex numbers whose modulus is 1. If a+bcosα+csinα=0, where α∈(0,π2), then

Let $a, b, c$ be three complex numbers whose modulus is $1$ . If $a + b cos α + c sin α = 0$ , where $α \in (0, \frac{π}{2})$ , then