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Definition of Vector

If a and b ar...

Question

If $\to a$ and $\to b$ are any two unit vectors, then the greatest positive integer in the range of $\frac{3 | \to a + \to b |}{2} + 2 | \to a - \to b |$ is

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Solution

Let the angle between $\to a$ and $\to b$ is $θ .$
We have, $| \to a | = | \to b | = 1$
and $| \to a + \to b | = \sqrt{1 + 1 + 2 cos θ} = 2 cos \frac{θ}{2}$ and
$| \to a - \to b | = \sqrt{1 + 1 - 2 cos θ} = 2 sin \frac{θ}{2}$
So, $\frac{3 | \to a + \to b |}{2} + 2 | \to a - \to b | = 3 cos \frac{θ}{2} + 4 sin \frac{θ}{2}$
So, maximum value $= \sqrt{3^{2} + 4^{2}} = 5$

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\to b and \to c

are two non-collinear unit vectors and

\to a

is any vector, then

(\to a \cdot \to b) \to b + (\to a \cdot \to c) \to c + \frac{\to a \cdot (\to b \times \to c)}{| \to b \times \to c |^{2}} (\to b \times \to c) =

\to b

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are two perpendicular unit vectors and

\to a

is any vector, then

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

\to a

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be two unit vectors if the vectors

\to c = \to a + 2 \to b

\to d = 2 \to a - 4 \to b

are perpendicular to each other. Then the angle between

\to a

\to b

\to a

\to b

are unit vectors, then the greatest value of

∣ ∣ ∣ \to a + \to b ∣ ∣ ∣ + ∣ ∣ ∣ \to a - \to b ∣ ∣ ∣

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| \to a | = 3

| \to b | = 1

| \to c | = 2

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