If e-1- e-2- e-3 -are vectors forming a reciprocal system to the vectors e1- e2- e3 then - e-1- e-2- e-2- -e1- e2- e4- is equal to

The correct option is A 1 The reciprocal system of vectors a, nbsp;b, nbsp;c nbsp;will be nbsp; b nbsp;×c[a nbsp;b nbsp; nbsp; ...

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Integration Using Substitution

If e'1, e'2...

Question

If ${\to e}_{1}^{'}, {\to e}_{2}^{'}, {¯ e}_{3}^{'}$ ,are vectors forming a reciprocal system to the vectors $_{1},_{2},_{3}$ then $[{\to e}_{1}^{'}, {\to e}_{2}^{'}, {\to e}_{2}^{'}]$ $[{\to e}_{1}, {\to e}_{2}, {\to e}_{4}]$ is equal to

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Similar questions

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

P (E)

E = E_{1} E_{2} E_{3} E_{4} E_{5}

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

P (E)

e_{1}

e_{2}

θ

s i n (\frac{θ}{2})

\frac{1}{2} | e 1 - e 2 |

If E1, E2, E3, ....., En are n elemantary events associated to a random experiment, then P(E1) + P(E2) + P(E3) + ........... + P(En) = ___

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