If the direction cosines of two lines are (fr | vedclass

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(C) Given,the direction cosines of two lines are $l_1 = (\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$ and $l_2 = (\frac{5}{13}, \frac{12}{13}, 0)$.The dire

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$\langle 41, 62, 13 \rangle$

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(C) Given,the direction cosines of two lines are $l_1 = (\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$ and $l_2 = (\frac{5}{13}, \frac{12}{13}, 0)$.The direction ratios of the line bisecting the angle between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are proportional to $\langle l_1+l_2, m_1+m_2, n_1+n_2 \rangle$.Substituting the values,the direction ratios are proportional to $\langle \frac{2}{3} + \frac{5}{13}, \frac{2}{3} + \frac{12}{13}, \frac{1}{3} + 0 \rangle$.Calculating the sum for each component:$\frac{2}{3} + \frac{5}{13} = \frac{26+15}{39} = \frac{41}{39}$$\frac{2}{3} + \frac{12}{13} = \frac{26+36}{39} = \frac{62}{39}$$\frac{1}{3} + 0 = \frac{13}{39}$Thus,the direction ratios are proportional to $\langle \frac{41}{39}, \frac{62}{39}, \frac{13}{39} \rangle$.Multiplying by $39$,we get the direction ratios as $\langle 41, 62, 13 \rangle$.

If the direction cosines of two lines are $(\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$ and $(\frac{5}{13}, \frac{12}{13}, 0)$,then identify the direction ratios of a line which is bisecting one of the angles between them.

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