Given- a- and b- are unit vector- and - be the angle between them-Then 1 - a-b-1 - a-b- -

The correct option is B tan ^2θ/2 a⃗·b⃗ = cosθ a⃗·b⃗ = cosθ 1 cosθ1 + cosθ = tan ^2θ/2 ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Given: a⃗ a...

Question

Given: $\to a$ and $\to b$ are unit vector, and $θ$ be the angle between them.
Then $\frac{1 - \to a . \to b}{1 + \to a . \to b} =$

{sin}^{2} \frac{θ}{2}

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{cos}^{2} \frac{θ}{2}

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{tan}^{2} \frac{θ}{2}

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{cot}^{2} \frac{θ}{2}

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Solution

The correct option is B ${tan}^{2} \frac{θ}{2}$
$\to a \cdot \to b = cos θ$
$\to a \cdot \to b = cos θ$
$\frac{1 - cos θ}{1 + cos θ} = {tan}^{2} \frac{θ}{2}$

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Given: →a and →b are unit vector, and θ be the angle between them.Then 1−→a.→b1+→a.→b=

The correct option is B tan2θ2→a⋅→b=cosθ→a⋅→b=cosθ1−cosθ1+cosθ=tan2θ2

Given: $\to a$ and $\to b$ are unit vector, and $θ$ be the angle between them.
Then $\frac{1 - \to a . \to b}{1 + \to a . \to b} =$

The correct option is B ${tan}^{2} \frac{θ}{2}$
$\to a \cdot \to b = cos θ$
$\to a \cdot \to b = cos θ$
$\frac{1 - cos θ}{1 + cos θ} = {tan}^{2} \frac{θ}{2}$