If vectors a - b and c are such that -a- - 1- -b- - 2 and -a -b-2-a -2b-2 - 20- then angle between vectors a and b is

Given, |a b|^2+|a +2b|^2 = 20 ⇒ nbsp; (|a|^2 +|b|^2 2a·b)+(|a|^2 +4|b|^2+4a·b) = 20 ⇒ (1+4 2a·b)+(1+4× 4+ 4a·b) = 20 ⇒ (5 2a·b) + (17+ 4a·b) = 20 ⇒22 +2a·b = ...

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Standard XII

Physics

Introduction

If vectors a ...

Question

If vectors $\to a, \to b$ and $\to c$ are such that $| \to a | = 1, | \to b | = 2$ and $| \to a - \to b |^{2} + | \to a + 2 \to b |^{2} = 20,$ then angle between vectors $\to a$ and $\to b$ is

\frac{π}{3}

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\frac{π}{4}

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\frac{π}{6}

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\frac{2 π}{3}

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Solution

The correct option is D $\frac{2 π}{3}$
Given, $| \to a - \to b |^{2} + | \to a + 2 \to b |^{2} = 20$
$\Rightarrow (| \to a |^{2} + | \to b |^{2} - 2 \to a \cdot \to b) + (| \to a |^{2} + 4 | \to b |^{2} + 4 \to a \cdot \to b) = 20$
$\Rightarrow (1 + 4 - 2 \to a \cdot \to b) + (1 + 4 \times 4 + 4 \to a \cdot \to b) = 20$
$\Rightarrow (5 - 2 \to a \cdot \to b) + (17 + 4 \to a \cdot \to b) = 20$
$\Rightarrow 22 + 2 \to a \cdot \to b = 20$
$\Rightarrow 2 \to a \cdot \to b = - 2$
$\Rightarrow \to a \cdot \to b = - 1$
$\Rightarrow | \to a | | \to b | cos θ = - 1$
$\Rightarrow 1 \times 2 cos θ = - 1$
$\Rightarrow cos θ = \frac{- 1}{2}$
$∴ θ = \frac{2 π}{3}$ (angle between vectors $\to a$ and $\to b$ )

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Introduction

Standard XII Physics

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If vectors →a,→b and →c are such that |→a|=1,|→b|=2 and |→a−→b|2+|→a+2→b|2=20, then angle between vectors →a and →b is

If vectors $\to a, \to b$ and $\to c$ are such that $| \to a | = 1, | \to b | = 2$ and $| \to a - \to b |^{2} + | \to a + 2 \to b |^{2} = 20,$ then angle between vectors $\to a$ and $\to b$ is