Let a-b-c- unit coplanar vectors b-c-3 then the maximum value of a-b-a-c- equals

The correct option is D √(3) We have, a .( b #160; + c) = | a|| b #160; + c|cosθ = 1 ×√(3)cosθ #160; = √(3)( max) for max ...

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Question

Let $\to a, \to b, \to c$ unit coplanar vectors $(\to b, \to c) = π / 3$ then the maximum value of $\to a . \to b + \to a . \to c$ equals

2

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\frac{3}{2}

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3

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\sqrt{3}

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\to a, \to b, \to c

\to a . \to b = \to a . \to c = 0

\to b

\to c

\frac{π}{3}

∣ ∣ [\begin{matrix} \to a & \to b & \to c \end{matrix}] ∣ ∣

\to a, \to b, \to c

\to a . \to b = 0 = \to a . \to c

\to b

\to c

π / 3

∣ ∣ ∣ \to a \times \to b - \to a \times \to c ∣ ∣ ∣

\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 C

\frac{\to A . \to B \times \to C}{\to C \times \to A . \to B} + \frac{\to B . \to A \times \to C}{\to C . \to A \times \to B}

| \to a | = 1, | \to b | = 2, | \to c | = 3

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

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

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