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Complex Numbers

Let c1=1,0,...

Question

Let $c_{1} = (1, 0, 0), c_{2} = (1, 1, 0), c_{3} = (1, 1, 1)$ , then the reciprocal of $c_{1} =$

A

^i +^j

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B

^i -^j

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C

^j -^k

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D

^k -^i

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Solution

The correct option is B $^i -^j$
Reciprocal of $C_{1} = \frac{C_{2} \times C_{3}}{[C_{1} C_{2} C_{3}]}$
Putting values we get
$= \frac{^i -^j}{1}$
$=^i -^j$

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Similar questions

^i \times (^j \times^k) +^j \times (^k \times^i) +^k \times (^i \times^j) =

\to a =^i +^j +^k

\to b =^i +^j -^k

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

\to d =^i -^j -^k

| (\to a \times \to b) . (\to c \times \to d) | =

\to a =^i +^j +^k; \to b =^i -^j +^k; \to c =^i +^j -^k

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

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

\to d =^i -^j -^k

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

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\to b =^i

\to c = c_{1}^i + c_{2}^j + c_{3}^k

c_{1} = 1

c_{2} = 2

c_{3}

which makes vector

(a, b)

\to c

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