If a- b- c- be unit vectors such that a-b-a-c-0 and the angle between b- and c- is -6 then a- -

The correct option is D ±2(b⃗×c⃗) a⃗ b⃗= a⃗ c⃗= 0 a⊥ to b⃗ amp; c⃗ a= k(b⃗×c⃗) | a⃗ |= #160; | k | | b⃗×c⃗ #160; | ...

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Theorems for Differentiability

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ be unit vectors such that $\to a \cdot \to b = \to a \cdot \to c = 0$ and the angle between $\to b$ and $\to c$ is $\frac{π}{6}$ then $\to a =$

\pm (\to a \times \to c)

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\pm 2 (\to a \times \to c)

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\pm (\to b \times \to c)

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\pm 2 (\to b \times \to c)

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Similar questions

\to a, \to b, \to c

\to a \cdot \to b = 0 = \to a \cdot \to c

\to b

\to c

\frac{π}{3}

| \to a \times \to b - \to a \times \to c |

\to a, \to b, \to c

\to a \cdot \to b = \to a \cdot \to c = 0

\to b

\to c

\frac{π}{3},

| \to a \times \to b - \to a \times \to c |

\to a, \to b

\to c, \to a \cdot \to b = \to a \cdot \to c

\to a \times \to b = \to a \times \to c

\to b = \to c

\to a, \to b

\to c

\to a \times (\to b \times \to c) = \frac{\sqrt{3}}{2} (\to b + \to c)

\to b

\to c

\to a

\to b

\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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