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Question Number 1016 by 123456 last updated on 14/May/15

[(\vec{x} \times \vec{a}) \times \vec{b}] \times \vec{x} = \vec{c}

Answered by rpatle69@gmail.com last updated on 14/May/15

[- \vec{b} \times (\vec{x} \times \vec{a})] \times \vec{x} = \vec{c}

\vec{x} \times [\vec{b} \times (\vec{x} \times \vec{a})] = \vec{c}

\vec{x} \times [(\vec{b} . \vec{a}) \vec{x} - (\vec{b} . \vec{x}) \vec{a}] = \vec{c}

Commented by prakash jain last updated on 17/May/15

\vec{x} \times (\vec{b} \cdot \vec{a}) \vec{x} - (\vec{b} . \vec{x}) (\vec{x} \times \vec{a}) = - (\vec{b} \cdot \vec{x}) (\vec{x} \times \vec{a})

Commented by 123456 last updated on 17/May/15

- (\vec{b} \cdot \vec{x}) (\vec{x} \times \vec{a}) \cdot \vec{a} = 0 = \vec{c} \cdot \vec{a}

- (\vec{b} \cdot \vec{x}) (\vec{x} \times \vec{a}) \cdot \vec{x} = 0 = \vec{c} \cdot \vec{x}

Commented by 123456 last updated on 17/May/15

\vec{x} = λ_{1} \vec{a} + λ_{2} (\vec{a} \times \vec{c})

for some λ_{1}, λ_{2} \in R

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