If a- - b- - 1 and a- - b- - j- - k- where a- - i- - j- - k- then b- is

The correct option is C i⃗ Let #160; b=b_xi+b_yj+b_zk #160; a̅.b̅ =b_x+b_y+b_z =1 ...(i) a̅×b̅ #160; =i(b_z b_y)+j(b_x ...

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Modulus of a Complex Number

If a⃗ . b⃗ ...

Question

If $\to a . \to b = 1$ and $\to a \times \to b = \to j - \to k$ , where $\to a = \to i + \to j + \to k$ , then $\to b$ is

\to i - \to j + \to k

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2 \to j - \to k

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\to i

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2 \to i

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Solution

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Similar questions

\to a = \to i + \to j + \to k, \to a = 2 \to i + \to k, \to c = 2 \to i + \to j + \to k; \to d = \to i + \to j + 2 \to k

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

\to a = \to i - 2 \to j - 3 \to k, \to b = 2 \to i + \to j - \to k, \to c = \to i + 3 \to j - 2 \to k

(\to a \times \to b) \times \to c

\to a

\to i

\to k

\to b

\to i

\to j

\to k

\to c

\to i

\to j

\to k

\to a

\to b

\to c

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

\to b

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

\to a . \to b = 3

\to b

\to i \times (\to a \times \to i) + \to j \times (\to a \times \to j) + \to k \times (\to a \times \to k)

\to i, \to j, \to k

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