Let a and b be two unit vectors inclined at an angle - then sin-2 is equal to

|â b̂|^2=|â|^2+|b̂|^2 2â·b̂ =1+1 2|â||b̂|cosθ =2(1 cosθ) =2(2sin^2θ2) |â b̂|^2=4sin^2θ2 |â b̂|= ± 2sinθ2 sinθ2 =|â b̂|2 nbsp; ...

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Standard XII

Mathematics

Unit Vectors

Let a and b b...

Question

Let $\to a$ and $\to b$ be two unit vectors inclined at an angle $θ,$ then $sin (\frac{θ}{2})$ is equal to

\frac{1}{2} ∣ ∣^a -^b ∣ ∣

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\frac{1}{2} ∣ ∣^a +^b ∣ ∣

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∣ ∣^a -^b ∣ ∣

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∣ ∣^a +^b ∣ ∣

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Solution

The correct option is A $\frac{1}{2} ∣ ∣^a -^b ∣ ∣$
${∣ ∣^a -^b ∣ ∣}^{2} = {|^a |}^{2} + {∣ ∣^b ∣ ∣}^{2} - 2^a \cdot^b$
$= 1 + 1 - 2 |^a | ∣ ∣^b ∣ ∣ cos θ$
$= 2 (1 - cos θ)$
$= 2 (2 {sin}^{2} \frac{θ}{2})$
${∣ ∣^a -^b ∣ ∣}^{2} = 4 {sin}^{2} \frac{θ}{2}$
$∣ ∣^a -^b ∣ ∣ = \pm 2 sin \frac{θ}{2}$
$sin \frac{θ}{2} = \frac{∣ ∣^a -^b ∣ ∣}{2}$

Suggest Corrections

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Unit Vectors

Standard XII Mathematics

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Let →a and →b be two unit vectors inclined at an angle θ, then sin(θ2) is equal to

The correct option is A 12∣∣^a−^b∣∣∣∣^a−^b∣∣2=|^a|2+∣∣^b∣∣2−2^a⋅^b =1+1−2|^a|∣∣^b∣∣cosθ =2(1−cosθ) =2(2sin2θ2) ∣∣^a−^b∣∣2=4sin2θ2 ∣∣^a−^b∣∣=±2sinθ2 sinθ2=∣∣^a−^b∣∣2

Let $\to a$ and $\to b$ be two unit vectors inclined at an angle $θ,$ then $sin (\frac{θ}{2})$ is equal to