If l_1, m_1, n_1 and l_2, m_2, n_2 are direct | vedclass

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(D) Let the unit vectors along $OA$ and $OB$ be $\vec{a} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}$ and $\vec{b} = l_2 \hat{i} + m_2 \hat{j} + n_2 \ha

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the unit vectors along $OA$ and $OB$ be $\vec{a} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}$ and $\vec{b} = l_2 \hat{i} + m_2 \hat{j} + n_2 \hat{k}$.Since $\vec{a}$ and $\vec{b}$ are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.The internal angular bisector of $\angle AOB$ is along the vector $\vec{v} = \vec{a} + \vec{b}$.$\vec{v} = (l_1+l_2) \hat{i} + (m_1+m_2) \hat{j} + (n_1+n_2) \hat{k}$.The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{(l_1+l_2)^2 + (m_1+m_2)^2 + (n_1+n_2)^2}$.$|\vec{v}|^2 = (l_1^2+m_1^2+n_1^2) + (l_2^2+m_2^2+n_2^2) + 2(l_1l_2 + m_1m_2 + n_1n_2)$.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:$|\vec{v}|^2 = 1 + 1 + 2 \cos 	heta = 2(1+\cos 	heta) = 4 \cos^2 \frac{	heta}{2}$.Thus,$|\vec{v}| = 2 \cos \frac{	heta}{2}$.The direction cosines of the internal bisector are the components of the unit vector $\frac{\vec{v}}{|\vec{v}|}$,which are:$\frac{l_1+l_2}{2 \cos \frac{	heta}{2}}, \frac{m_1+m_2}{2 \cos \frac{	heta}{2}}, \frac{n_1+n_2}{2 \cos \frac{	heta}{2}}$.

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of $OA$ and $OB$ such that $\angle AOB = \theta$,where $O$ is the origin,then the direction cosines of the internal angular bisector of $\angle AOB$ are

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