If u-a-b- - v-a-b- and - a- - - b- -2 then - u-v- - is equal to

The correct option is B 2√( ( 16 ( a⃗.b⃗ )^2 )) u⃗×v⃗ = (a⃗ b⃗ )× (a⃗ +b⃗ ) #160;= 0+ a⃗×b⃗ b⃗×a⃗ 0 Now, b⃗×a⃗ = a⃗×b⃗ Thus ...

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Definition of Functions

If u⃗=a⃗-b⃗...

Question

If $\to u = \to a - \to b; \to v = \to a + \to b$ and $| \to a | = ∣ ∣ \to b ∣ ∣ = 2$ then $| \to u \times \to v |$ is equal to

\sqrt{2 (16 - {(\to a . \to b)}^{2})}

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2 \sqrt{(16 - {(\to a . \to b)}^{2})}

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2 \sqrt{(4 - {(\to a . \to b)}^{2})}

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[4 - (\to a . \to b)^{2}]

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Solution

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\to u = \to a - \to b, \to v = \to a + \to b

| \to a | = | \to b | = 2

| \to u \times \to v |

$I f \to u = \to a - \to b a n d \to v = \to a + \to b a n d | \to a | = | \to b | = 2, t h e n | \to u \times \to v | =$

\frac{(\to a \times \to b)^{2} + (\to a . \to b)^{2}}{2 a^{2} b^{2}}

| \to a | = a, | \to b | = b

\to a

\to b

\to u = \to a - (\to a . \to b) \to b

\to v = \to a \times \to b,

| \to v |

\to u = \to a \times (\to b \times \to c) + \to b \times (\to c \times \to a) + \to c \times (\to a \times \to b)

\to u =

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