The vectors a- and b- have magnitudes 2 and 2 -2 respectively- It is found that a- - b-a-b- Then the value of -a-b-a-b- will b

The correct option is B a . nbsp;b= |a × nbsp; nbsp;b| ab cos θ = ab sin θ , θ = nbsp; nbsp;45^∘ ∴ |a + b/a b| = nbsp;√(a^2 + b^2+ 2ab cos θ/a ...

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Range of Trigonometric Expressions

The vectors a...

Question

The vectors $\to a$ and $\to b$ have magnitudes 2 and 2 $\sqrt{2}$ respectively. It is found that $\to a$ . $\to b$ = | $\to a$ $\times$ $\to b$ |. Then the value of | $\frac{\to a + \to b}{\to a - \to b}$ | will be

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Solution

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\to a

\to b

\frac{{∣ ∣ \to a + \to b ∣ ∣}^{2} - {∣ ∣ \to a - \to b ∣ ∣}^{2}}{{∣ ∣ \to a + \to b ∣ ∣}^{2} + {∣ ∣ \to a - \to b ∣ ∣}^{2}}

\to a, \to b, \to c

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 2, | \to c | = 4

\to a + \to b + \to c = \to 0

\to a . \to b + 3 \to b . \to c + 3 \to c . \to a

\to a, \to b, \to c

\to a . \to b = \to a . \to c, \to a \times \to b = \to a \times \to c, \to a \neq 0

\to b = \to c

\to a

\to b

| \to a \times \to b |^{2} = | \to a |^{2} | \to b |^{2} - | \to a . \to b |^{2}

| \to a + \to b | = | \to a - \to b |

(\to a, \to b) = \frac{π}{2}

\to a, \to b, \to a + \to b

(\to a, \to b) = \frac{2 π}{3}

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