1-If-a-a-1i-a-2j-a-3k-and-b-b-1i-b-2j-b-3k-show-that-i-a-b-determinant-i-j-k-a-1-a-2-a-3-b-1-b-2-b-3-ii-a-b-a-1-b-1-a-2-b-2-a-3-b-3-2-If-a-

Question Number 44848 by pieroo last updated on 05/Oct/18

(1) If \vec{a} = a_{1 i} + a_{2 j} + a_{3 k} and \vec{b} = b_{1 i} + b_{2 j} + b_{3 k} show that

i . \vec{a} \times \vec{b} = | \begin{matrix} i & j & k \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix} |

ii . \vec{a} ∙ \vec{b} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

(2) If \vec{a} = a_{1 i} + a_{2 j} + a_{3 k}, \vec{b} = b_{1 i} + b_{2 j} + b_{3 k} and

\vec{c} = c_{1 i} + c_{2 j} + c_{3 k}, show that

i . \vec{a} ∙ (\vec{b} \times \vec{c}) = | \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} |

ii . a ∙ (\vec{b} \times \vec{c}) = \vec{b} (\vec{a} ∙ \vec{c}) - \vec{c} (\vec{a} ∙ \vec{b})

iii . (\vec{a} \times \vec{b}) \times \vec{c} = \vec{b} (\vec{a} ∙ \vec{c}) - \vec{a} (\vec{b} ∙ \vec{c})

iv . \vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = 0

Commented by pieroo last updated on 05/Oct/18

please I need help with the above questions

please I need help with the above questions

Commented by pieroo last updated on 06/Oct/18

still waiting for some help, please

still waiting for some help, please

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