Lf unit vectors A- and B- are inclined at an angle - then -A-B-

The correct option is B 2sin(θ/2) |A B| = √(A^2+B^2 2A · B cosθ) but |A| = |B| = 1 #160;Hence, #160; |A B| #160;=√(2 2cosθ) =√(2(1 c ...

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Standard XII

Physics

Vector Component

lf unit vecto...

Question

lf unit vectors $\to A$ and $\to B$ are inclined at an angle $θ$ , then $| \to A - \to B |$ :

2 cos (\frac{θ}{2})

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2 sin (\frac{θ}{2})

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Solution

The correct option is B $2 sin (\frac{θ}{2})$
$| \to A - \to B | = \sqrt{A^{2} + B^{2} - 2 A \cdot B cos θ}$
but $| A | = | B | = 1$
Hence, $| \to A - \to B | = \sqrt{2 - 2 cos θ} = \sqrt{2 (1 - cos θ)} = \sqrt{4 {sin}^{2} (\frac{θ}{2})}$ = $2 sin (\frac{θ}{2})$

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lf unit vectors →A and →B are inclined at an angle θ, then |→A−→B|:

The correct option is B 2sin(θ2)|→A−→B|=√A2+B2−2A⋅B cosθbut |A|=|B|=1 Hence, |→A−→B|=√2−2cosθ=√2(1−cosθ)=√4sin2(θ2)=2sin(θ2)

lf unit vectors $\to A$ and $\to B$ are inclined at an angle $θ$ , then $| \to A - \to B |$ :

The correct option is B $2 sin (\frac{θ}{2})$
$| \to A - \to B | = \sqrt{A^{2} + B^{2} - 2 A \cdot B cos θ}$
but $| A | = | B | = 1$
Hence, $| \to A - \to B | = \sqrt{2 - 2 cos θ} = \sqrt{2 (1 - cos θ)} = \sqrt{4 {sin}^{2} (\frac{θ}{2})}$ = $2 sin (\frac{θ}{2})$