If (alpha, beta, gamma) are the direction cos | vedclass

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(B) Let the direction ratios of the two lines be $\vec{a} = (2, 2, 1)$ and $\vec{b} = (2, -1, -2)$.First,we normalize these vectors to find the unit v

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$2$

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(B) Let the direction ratios of the two lines be $\vec{a} = (2, 2, 1)$ and $\vec{b} = (2, -1, -2)$.First,we normalize these vectors to find the unit vectors (direction cosines):$|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$,so $\hat{a} = (\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$.$|\vec{b}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3$,so $\hat{b} = (\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3})$.The angular bisector direction is given by the vector $\vec{v} = \hat{a} + \hat{b}$ or $\vec{v} = \hat{a} - \hat{b}$.Case $1$: $\vec{v}_1 = (\frac{2}{3} + \frac{2}{3}, \frac{2}{3} - \frac{1}{3}, \frac{1}{3} - \frac{2}{3}) = (\frac{4}{3}, \frac{1}{3}, -\frac{1}{3})$.Magnitude $|\vec{v}_1| = \sqrt{(\frac{4}{3})^2 + (\frac{1}{3})^2 + (-\frac{1}{3})^2} = \sqrt{\frac{16+1+1}{9}} = \sqrt{2}$.Direction cosines $(\alpha, \beta, \gamma) = (\frac{4}{3\sqrt{2}}, \frac{1}{3\sqrt{2}}, -\frac{1}{3\sqrt{2}})$.Then $(\alpha + \beta + \gamma)^2 = (\frac{4+1-1}{3\sqrt{2}})^2 = (\frac{4}{3\sqrt{2}})^2 = \frac{16}{18} = \frac{8}{9}$.Case $2$: $\vec{v}_2 = (\frac{2}{3} - \frac{2}{3}, \frac{2}{3} + \frac{1}{3}, \frac{1}{3} + \frac{2}{3}) = (0, 1, 1)$.Magnitude $|\vec{v}_2| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.Direction cosines $(\alpha, \beta, \gamma) = (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.Then $(\alpha + \beta + \gamma)^2 = (0 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 = (\frac{2}{\sqrt{2}})^2 = (\sqrt{2})^2 = 2$.

If $(\alpha, \beta, \gamma)$ are the direction cosines of an angular bisector of two lines whose direction ratios are $(2, 2, 1)$ and $(2, -1, -2)$,then $(\alpha + \beta + \gamma)^2 = $

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