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Angle between Two Vectors

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Question

If $\to a$ and $\to b$ are two vectors of magnitude $1$ inclined at $120^{\circ}$ , then find the angle between $\to b$ and $\to b - \to a$ .

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Solution

$\to a \cdot \to b = | \to a | | \to b | cos 120^{\circ}$

$= 1 \times 1 \times cos (180 - 60)$

$= - cos 60^{\circ} = - \frac{1}{2}$

Let the angle between $\to b$ and $\to b - \to a$ be $θ$

Then,

$\begin{matrix} \to b \cdot (\to b - \to a) = ∣ ∣ \to b ∣ ∣ ∣ ∣ \to b - \to a ∣ ∣ cos θ \to b . \to b - \to a . \to b = 1 \times ∣ ∣ \to b - \to a ∣ ∣ cos θ (\to a . \to b = \to b . \to a) {∣ ∣ \to b ∣ ∣}^{2} - \to a . \to b = ∣ ∣ \to b - \to a ∣ ∣ cos θ 1 + \frac{1}{2} = ∣ ∣ \to b - \to a ∣ ∣ cos θ ∣ ∣ \to b - \to a ∣ ∣ cos θ = \frac{3}{2} - - - - (1) \end{matrix}$

Now,

$\begin{matrix} {∣ ∣ \to b - \to a ∣ ∣}^{2} = (\to b - \to a) (\to b - \to a) = \to b . \to b - \to b . \to a - \to a . \to b + \to a . \to a = {∣ ∣ \to b ∣ ∣}^{2} - 2 \to a . \to b + {| \to a |}^{2} {∣ ∣ \to b - \to a ∣ ∣}^{2} = 1 - 2 \times \frac{- 1}{2} + 1 = 3 ∣ ∣ \to b - \to a ∣ ∣ = \sqrt{3} \end{matrix}$

Puttimg the values of $| \to b - \to a |$ in $(1)$

$| \to b - \to a | cos θ = \frac{3}{2}$

$\sqrt{3} cos θ = \frac{3}{2}$

$cos θ = \frac{\sqrt{3}}{2}$

$θ = 30^{\circ}$

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