The angle in degree between two planar vectorsa-3-2-1-2- and b- - -3-2-1-2- is

A⃗.b⃗= ( √(3)î/2+1/2ĵ ). ( √(3)î/2+ĵ/2 ) = 3/4+1/4= 1/2 | a⃗ |=1 nbsp; amp; nbsp;| b⃗ |=1 nbsp; nbsp; So θ= cos^ 1 ( a⃗.b⃗/ | a⃗ | | b⃗ | )=cos^ 1 ...

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Byju's Answer

Standard IX

Mathematics

Dot Product of Two Vectors

The angle in ...

Question

The angle (in degree ) between two planar vectors
$\to a = \frac{\sqrt{3}}{2}^i + \frac{1}{2}^j$ and $\to b$ = $\frac{- \sqrt{3}}{2}^i + \frac{1}{2}^j$ is

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120

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Solution

The correct option is D 120
$\to a . \to b = (\frac{\sqrt{3}^i}{2} + \frac{1}{2}^j) . (\frac{- \sqrt{3}^i}{2} + \frac{^j}{2})$
= $\frac{- 3}{4} + \frac{1}{4} = \frac{- 1}{2}$
$| \to a | = 1$ & $∣ ∣ \to b ∣ ∣ = 1$

So
$θ = c o s^{- 1} (\frac{\to a . \to b}{| \to a | ∣ ∣ \to b ∣ ∣}) = c o s^{- 1} (\frac{- 1}{2}) = 120^{\circ}$

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The angle (in degree ) between two planar vectors →a=√32^i+12^j and →b = −√32^i+12^j is

The correct option is D 120→a.→b=(√3^i2+12^j).(−√3^i2+^j2) =−34+14=−12 |→a|=1 & ∣∣→b∣∣=1 So θ=cos−1(→a.→b|→a|∣∣→b∣∣)=cos−1(−12)=120∘

The angle (in degree ) between two planar vectors
$\to a = \frac{\sqrt{3}}{2}^i + \frac{1}{2}^j$ and $\to b$ = $\frac{- \sqrt{3}}{2}^i + \frac{1}{2}^j$ is