If the direction cosines of two lines incline | vedclass

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(B) Let $OA$ and $OB$ be two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively. Taking $OA = OB = 1$,the coordinates o

Vedclass · Accepted Answer

$\frac{l_1 + l_2}{2 \cos(	heta/2)}, \frac{m_1 + m_2}{2 \cos(	heta/2)}, \frac{n_1 + n_2}{2 \cos(	heta/2)}$

Vedclass · Answer

(B) Let $OA$ and $OB$ be two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ respectively. Taking $OA = OB = 1$,the coordinates of $A$ and $B$ are $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.Let $OC$ be the internal bisector of $\angle AOB$. Since $OA = OB$,$OC$ is the median of $	riangle OAB$,and $C$ is the midpoint of $AB$.The coordinates of $C$ are $\left( \frac{l_1 + l_2}{2}, \frac{m_1 + m_2}{2}, \frac{n_1 + n_2}{2} ight)$.The direction ratios of $OC$ are $\left( \frac{l_1 + l_2}{2}, \frac{m_1 + m_2}{2}, \frac{n_1 + n_2}{2} ight)$.Now,the length $OC = \sqrt{\left( \frac{l_1 + l_2}{2} ight)^2 + \left( \frac{m_1 + m_2}{2} ight)^2 + \left( \frac{n_1 + n_2}{2} ight)^2}$.$OC = \frac{1}{2} \sqrt{(l_1^2 + m_1^2 + n_1^2) + (l_2^2 + m_2^2 + n_2^2) + 2(l_1l_2 + m_1m_2 + n_1n_2)}$.Since $l_i^2 + m_i^2 + n_i^2 = 1$ and $l_1l_2 + m_1m_2 + n_1n_2 = \cos 	heta$,we have:$OC = \frac{1}{2} \sqrt{1 + 1 + 2 \cos 	heta} = \frac{1}{2} \sqrt{2(1 + \cos 	heta)} = \frac{1}{2} \sqrt{2(2 \cos^2(	heta/2))} = \cos(	heta/2)$.The direction cosines of $OC$ are obtained by dividing the direction ratios by the length $OC$:$\frac{l_1 + l_2}{2 \cos(	heta/2)}, \frac{m_1 + m_2}{2 \cos(	heta/2)}, \frac{n_1 + n_2}{2 \cos(	heta/2)}$.

If the direction cosines of two lines inclined at an angle $\theta$ are $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,then the direction cosines of the internal bisector of the angle between the lines are:

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