If a-b-c- are unit vectors such that a-b-c-0- then a-b-b-c-c-a-

The correct option is B 32 a⃗ + b⃗ + c⃗ = 0 ⇒ (a⃗ + b⃗ + c⃗).(a⃗ + b⃗ + c⃗) = 0 ⇒ |a⃗|^2 + a⃗.b⃗ + a⃗.c⃗ + b⃗.a⃗ + |b⃗|^2 + b⃗.c⃗ + c⃗.a⃗ ...

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Vector Component

If a⃗,b⃗,c⃗...

Question

If $\to a, \to b, \to c$ are unit vectors such that $\to a + \to b + \to c = \to 0$ , then $\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a =$ _____

\frac{3}{2}

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- \frac{3}{2}

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\frac{2}{3}

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\frac{1}{3}

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Solution

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\to a, \to b, \to c

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

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

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

| \to a | = 3, | \to b | = 5

| \to c | = 7

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 1, | \to c | = 4

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

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

\to a, \to b, \to c

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

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

\to a, \to b, \to c

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

(\to a, \to b) = \frac{π}{3}

∣ ∣ \to a \times \to b ∣ ∣ + ∣ ∣ \to b \times \to c ∣ ∣ + | \to c \times \to a | =

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