Let a and b be two non -null vectors such that -a- b- - -a - 2b- Then the value of -a-b- may be

|a+ b| =|a 2b| Squaring on both sides, a^2 + b^2 + 2ab cosθ = a^2 + 4b^2 4ab cosθ cosθ = b/2a≤ 1 ∴ b ≤ 2a ab≥1/2 Hence, the possible answers are 1 and 2 acco ...

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Byju's Answer

Standard XIII

Physics

Vector Addition

Let a and b b...

Question

Let $\to a$ and $\to b$ be two non -null vectors such that $| \to a + \to b | = | \to a - 2 \to b |$ . Then the value of $\frac{| \to a |}{| \to b |}$ may be

\frac{1}{4}

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\frac{1}{8}

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Solution

The correct option is D 2
$| \to a + \to b | = | \to a - 2 \to b |$
Squaring on both sides,
$a^{2} + b^{2} + 2 a b cos θ = a^{2} + 4 b^{2} - 4 a b cos θ$
$cos θ = \frac{b}{2 a} \leq 1$
$∴ b \leq 2 a$
$\frac{a}{b} \geq \frac{1}{2}$
Hence, the possible answers are 1 and 2 according to options.

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Let →a and →b be two non -null vectors such that |→a+→b|=|→a−2→b|. Then the value of |→a||→b| may be

The correct option is D 2|→a+→b|=|→a−2→b| Squaring on both sides, a2+b2+2abcosθ=a2+4b2−4abcosθ cosθ=b2a≤1 ∴b≤2a ab≥12 Hence, the possible answers are 1 and 2 according to options.

Let $\to a$ and $\to b$ be two non -null vectors such that $| \to a + \to b | = | \to a - 2 \to b |$ . Then the value of $\frac{| \to a |}{| \to b |}$ may be

The correct option is D 2
$| \to a + \to b | = | \to a - 2 \to b |$
Squaring on both sides,
$a^{2} + b^{2} + 2 a b cos θ = a^{2} + 4 b^{2} - 4 a b cos θ$
$cos θ = \frac{b}{2 a} \leq 1$
$∴ b \leq 2 a$
$\frac{a}{b} \geq \frac{1}{2}$
Hence, the possible answers are 1 and 2 according to options.