The value of i- a-i- - j- a-j- - k- a-k- is where i- j- k- are unit vectors

The correct option is B 2 a⃗ i × (a̅× i) = (i . i) a̅ (i. a̅) i = a̅ (i . a̅)i so i × (a̅× i) + j × (a̅× j) + k × (a̅× k) = a̅ (i . a ...

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Vectors and Its Types

The value of ...

Question

The value of $\to i \times (\to a \times \to i) + \to j \times (\to a \times \to j) + \to k \times (\to a \times \to k)$ is (where $\to i, \to j, \to k$ are unit vectors)

\to a

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2 \to a

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- \to a

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\to a = \to i + 2 \to j + \to k, \to b = \to i - \to j + \to k

\to c = \to i + \to j - \to k

\to a

\to b

\frac{1}{\sqrt{3}} o n \to c

\to a, \to b, \to c

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

\to a

\to b & \to c

{\to v}_{1}, {\to v}_{2}

{\to v}_{3}

{\to v}_{1} = {\to v}_{2} - {\to v}_{3}

{\to v}_{1} = (\to a \times ˆ i) \times ˆ i

{\to v}_{2} = (\to a \times ˆ j) \times ˆ j

{\to v}_{3} = (\to a \times ˆ k) \times ˆ k

\to a

\to a, \to b, \to c

\to d

(\to a \times \to b) \times (\to c \times \to d) = [\to a \to c \to d] \to b - [\to b \to c \to d] \to a

\to a = i + j - k, \to b = 1 - j + k, \to c

\to c . \to a = 0, [\to c \to a \to b] = 0

\to a

\to c

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