If vec{a}, vec{b}, vec{c} are three non-zero, | vedclass

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(B) We check the orthogonality of the set $\{\vec{a}, \vec{b_1}, \vec{c_2}\}$.First,$\vec{a} \cdot \vec{b_1} = \vec{a} \cdot \left( \vec{b} - \frac{\v

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$\{\vec{a}, \vec{b_1}, \vec{c_2}\}$

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(B) We check the orthogonality of the set $\{\vec{a}, \vec{b_1}, \vec{c_2}\}$.First,$\vec{a} \cdot \vec{b_1} = \vec{a} \cdot \left( \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} \right) = \vec{a} \cdot \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} |\vec{a}|^2 = \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} = 0$.Thus,$\vec{a}$ and $\vec{b_1}$ are orthogonal.Next,$\vec{a} \cdot \vec{c_2} = \vec{a} \cdot \left( \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1} \right) = \vec{a} \cdot \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} |\vec{a}|^2 - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2} (\vec{a} \cdot \vec{b_1}) = \vec{a} \cdot \vec{c} - \vec{c} \cdot \vec{a} - 0 = 0$.Thus,$\vec{a}$ and $\vec{c_2}$ are orthogonal.Finally,$\vec{b_1} \cdot \vec{c_2} = \vec{b_1} \cdot \left( \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1} \right) = \vec{b_1} \cdot \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2} (\vec{b_1} \cdot \vec{a}) - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2} |\vec{b_1}|^2 = \vec{b_1} \cdot \vec{c} - 0 - \vec{c} \cdot \vec{b_1} = 0$.Thus,$\vec{b_1}$ and $\vec{c_2}$ are orthogonal.Since all pairs are orthogonal,$\{\vec{a}, \vec{b_1}, \vec{c_2}\}$ is a set of mutually orthogonal vectors.

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